blog #neewollahyppah

happy halloween everyone. so nyways last week we had sub with review for the latter part so i dont remember much, sorry...

Devin's Reflection

Mean Value Theorem
If f is continuous on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists a number c in (a,b) such that

f'(c) = f(b) - f(a)/b-a

The "mean" in the Mean value Theorem refers to the mean rate of change of f in the interval [a,b]. In the equation y = f(1) - f(2)/1-2(x-2) - f(2)
Let g(x) = f(x) - y
So it would be f(x) - the equation
By evaluating g at a and b, you can see that g(a) = 0 = g(b). Because f is continuous on [a,b], it follows that g is also continuous on [a,b]. Furthermore, because f is differentiable, g is also differentiable, and you can apply Rolle's Theorem to the function g. So, there exists a number c in (a,b) such that g'(c) = 0, which implies that function g. So, there exists a number c (a,b) such that g'(c) = 0, which implies that 0 = g'(c)
= f'(c) - f(1) - f(2)/1-2
So, there exists a number c in (a,b) such that
f'(c) = f(1) - f(2)/1-2

halloween blog! :D

HAPPY HALLOWEEN YOUS GAIS! haha.

a brief overview of what we learned last week:

Rolle's theorem - used to find max and min on a given interval; check differentiability, check continuity. if it is uncontinuous and nondifferentiable, rolle's theorem cannot be applied; if it is continuous and differentiable, continue by plugging in intervals into the original problem, then take the derivative and set = to 0. solve for x

Mean Value Theorem - check differentiability, check continuity, check slope, take derivative and set = 0, solve for x.

First Derivative Test - take derivative (duh) and set = 0, solve for x, set up intervals: (-infinity, ) u ( , infinity) plug in value on interval into the derivative

Extreme values - maxes and minimums; absolute max is the highest point, and absolute min is the lowest point.

halloween blog

Happy Halloween to all you ghouls and goblins

lol, i had to say that

anyways, I don't really remember much about last week's math classes, but I do remember something like the First Derivative Test from last week, I think, but I don't understand it, like the rest of the math this year

taylor Rodriguez blog #2

This week all we did was review so I will reiterate on all of chapter three which we have learned so far so that lessons 1-3 are all in one place in case there is a need to print, study, refer to etc...

CHAPTER 3 LESSON 1
Extreme Values are defined as maximums and minimums
Absolute Max is defined as the highest point on an interval
** notice that absolute max is the highest point on a given interval not the highest point on the graph
Absolute min is defined as the lowest point on an interval
**notice that absolute min is the lowest point of a given interval not the lowest point on the graph
Relative Max (also referred to as local max) is defined as any max NOT on an interval
Relative Min (also referred to a local min) is defined as any min NOT on an interval
Critical Numbers are defined as any max or min typically referred to as x=c
When trying to solve problems concerning a critical number, extreme value, max, min, or horizontal tangents take the derivative and set equal to zero.

Example #1:
Find the critical values of f(x)= 9(x^2-3)/x^3
9x^2-27/x^3
Take the derivative
X^3(18x)-[(9x^2-27)(3x^2)]/ x^6
18x^4-[27x^4-81x^2]/x^6
18x^4-27x^4+81x^2/x^6
-9x^4+81x^2/ x^6
9x^2(-x^2+9)/X^6
9(-x^2+9)/x^4
Set equal to zero and solve
((when setting a fraction equal to zero you are really only setting what is in the numerator equal to zero))
9(-x^2+9)=0
-9x^2+81=0
-9x^2=-81
X^2=9
X=+/- 3

Example #2:
Find the extrema of f(x)= 3x^4-4x^3 on the interval [-1,2]
4(3)x^3-3(4)x^2
12x^3-12x^2=0
12x^2(x-1)=0
X=0,1

To determine max or min on an interval plug into original and the biggest result will be the max
and the smallest result will be the min
f(-1)= 3(-1)^4-4(-1)^3=7
f(0)=3(0)^4-4(0)^3=0
f(1)= 3(1)^4 – 4(1)^3= -1 ((smallest so absolute min))
f(2)= 3(2)^4-4(2)^3= 16 ((largest so absolute max))

CHAPTER 3 LESSON 2
Rolle’s Thorem:
Rolle’s Thorem is used to discover whether there is a max or a min on the given interval.
Steps:
Check differentiability
Check for continuity
If the problem is not differentiable or if the problem is not continuous Rolle’s thorem cannot be applied.
If the problem is differentiable and is continuous then you will continue to work the problem by:
Test that the y values match by plugging interval numbers into the original problem
Take the derivative and set equal to zero
Solve for x
**Remember f’(c)= 0 must be read as max or min and therefore you must use rolle’s thorem
Example:
Let F(x)=x^4-2x^2 find all values of c in the interval [-2,2] such that f’(c)=0
Check differentiability
Differentiable? Yes.
Check for continuity
Continuous? Yes
Test that the y values match by plugging interval numbers into the original problem
F(-2)= -2^4-2(-2)^2
16-2(4)
16-8
=8
F(2)= 2^4-2(2)^2
16-2(4)
16-8
=8
Therefore, yes
Take the derivative and set equal to zero
4x^3-4x=0

Solve for x
4x(x^2-1)=0
X=0 X=-1 X=1
Therefore by Rolle’s thorem c= 0, -1, 1
By Rolle’s theorm the function is continuous, differentiable, and f(1)=f(2) thus there must be one max or min on the interval [-2, 2] at f’(x)=0

Mean Value Thorem: The mean value theorem is used to prove that on the interval there must be some x value where the derivative equals the slope between the points of the interval
Steps:
Check Differentiability
Check Continuity
If the problem is not differentiable or if the problem is not continuous the Mean Value theorem cannot be applied.
If the problem is differentiable and is continuous then you will continue to work the problem by:
Check Slope (((( f’ (C)= f(b)-f(a)/b-a))))
Take the derivative and set equal to zero
Solve for x
Example: Given f(x)= 5-(4/x) find all values of c in the open interval (1,4) such that f’(c)= f(4)-f(1)/ 4-1
Check Differentiability
Differentiable? Yes
Check Continuity
Continuous? Yes
Check Slope (((( f’ (C)= f(b)-f(a)/b-a))))
4-1/4-1 = 1

Take the derivative and set equal to slope
-4(-1)x^-2
4x^-2
4/x^2= 1
Solve for x
X=+/- 2
Therefore by the mean value theorem, the function is continuous and differentiable on the interval therefore there is a value c=2 where the derivative = slope on the interval (1,4)

CHAPTER 3 LESSON 3
The first derivative test:
Used to determine whether it is specifically a max or a min on the interval
KEY WORDS TO RECOGNIZE WHEN TO USE THE FIRST DERIVATIVE TEST
Increasing, decreasing, max, min, horizontal tangent, change direction
Steps:
Take derivative and set equal to zero
Solve for x
Set up intervals (-infinity, pt)U(pt, infinity)
Plug in value on the interval into the derivative ((if you get a positive number the function is increasing, If you get a negative number the function is decreasing)) (( A max is increasing then decreasing, A min is decreasing then increasing))
Example:
Find the open intervals on which f(x)= x^3-3/2x^2 is increasing or decreasing
Take derivative and set equal to zero
3x^2-3x=0
Solve for x
3x(x-1)=0
X=0,1
Set up intervals (-infinity, pt)U(pt, infinity)
(negative infinity,0)U(0,1)U(1,infinity)
Plug in value on the interval into the derivative
F’(-1)= 3(-1)^2-3(-1)=6
Positive therefore increasing
F’(1/2)= 3(1/2)^2-3(1/2)= -3/4
Negative therefore decreasing
F’(2)= 3(2)^2-3(2)= 6
Positive therefore increasing
Therefore increasing: (negative infinity, 0) U (1, infinity)
Decreasing: (0,1)
Max: at 0
Min: at 1

Blog 10

Happy Halloween! :)

Anyways, We just finished chapter three which was all the guidelines stuff. It was pretty easy but just like every other chapter they had parts that were very confusing. For me the confusing stuff was the problems that dealt with trig. I still don’t understand the whole quadrant thing; I just forgot it from last year. But something I do remember was the two theorems and I’ll explain them since I didn’t get to last blog. So, the two theorems are Rolle’s Theorem and Mean Value theorem. The confusing part is that they both deal with max’s and min’s. Here are some hints to tell them apart: Rolle’s Theorem only applies to questions on an interval, and it tells you that there is at least one max or min, however there could be more than one. Now to find out if the number is on the interval, you’re going to take the derivative and set it equal to zero. If the number is in the interval then you would plug it back into the original equation to see if it is a max or min. **Typically Rolle’s Theorem uses x-intercepts. The Mean Value Theorem deals with slope. You set the slope equal to the derivative and solve to find the number in-between.

EXAMPLE OF ROLLE’S THEOREM:
Ex. 1: Find the 2x-intercepts of f(x)=x^2-3x+2 and show that f’(x)=0 at some point between the 2x-intercepts.
*Check if continuous-It’s continuous
*Check if differentiable-Yes
*Solve for 0 to find interval
X^2-3x+2=0
(x-2)(x-1)=0
X=1,2 [1,2]

EXAMPLE OF MEAN VALUE THEOREM:
Ex. 2: Given f(x)=5-(4/x) find all values of c in the open interval (1,4) such that f’(c)=f(4)-f(1)/4-1
*Continuous
*Differentiable
*Set slope=derivative
4=x^2
X=+/- 2

Blog #10

First off, HAPPY HALLOWEEN!!

Anyway, this week we took the Chapter 3 test, so all we did was review for the test. Because of that, for this blog I’ll just go over the First Derivative Test.

Step 1: Take the derivative.
Step 2: Set it equal to zero.
Step 3: Solve for x.
Step 4: Set up intervals. (-infinity, _ ) u ( _, infinity)
Step 5: Plug a value on the interval into the derivative. If you get a positive number, the function is increasing. If you get a negative number, the function is decreasing.
Step 6: Figure out the max and min. (A max is increasing then decreasing and a min is decreasing then increasing.)

Ex. Find the open intervals on which f(x) = x^3 – 3/2x^2 is increasing or decreasing.
1) 3x^2 – 3x = 0
2) 3x (x – 1) = 0
3) x = 0, 1
4) (-infinity, 0) u (0, 1) u (1, infinity)
5) f’(-1) = 3(-1)^2 – 3(-1) = 6 increasing
f’(1/2) = 3(1/2)^2 – 3(1/2) = -3/4 decreasing
f’(2) = 3(2)^2 – 3(2) = 6 increasing
6) increasing: (-infinity, 0) u (1, infinity)
decreasing: (0, 1)

*Note: If they ask for absolute extrema, you have to plug back into the original equation to find the y-values in order to find the critical points and end points.

Halloween Blog of the Dead - topsgolbredrum

This week went by pretty fast, although not one of the best weeks. Pretty tired, almost killed myself nd my parents in a car crash yesterday, sleepin at the wheel, tsk tsk. ANWAYS, to the math, i guess. We didn't learn much this week, except for continuing blah blah in chapter 3 and goin back to chapters 1 and 2. I guess i'll explain how to do some more fricking derivative, using the Mean Value Theorem.

First, you have to check if the function is continuous or not. If it is, move to step 2.

Second, you have to check if the function is differentiable or not. If it is, move to step 3.

Third, you have to do something with the slope.

Use the function

f ' (c) = (f(b) - f(a))/(b-a)

It means on an interval [a,b]. There must be some x-value where the derivative equals the slope between a & b.

slope of a secant line = slope of a tangent line between a & b.

Example:

Given f(x) = 5 - (4/x), find all the values of c in the open interval (1,4) such that

f ' (c) = (f(4) - f(1))/(4-1)

1. continuous? yes

2. differentiable? yes

4-1/4-1 = 3

5 - 4x^-1

4/x^2 = 1

4 = x^2

x = plus or minus 2

c = plus or minus 2

By the mean value theorem, the function is continuous and differentiable, therefore the same value c=2 where f ' (x) = the slope between the open interval (1,4).

10/31/10

Soooo, this week was NOT a good week for me at all..Basically from what I remember, we didn't learn anything new. We just practiced with the concepts that we recently learned in Chapter 3, and of course some things from Chapters 1 and 2 were thrown in there too. So here are a few examples:

Ex. 1) Find all critical numbers of the function g(x)=x^4-26x^2
*key words in this problem are "critical numbers"..that tells you that all you have to do is take the derivative of the function and set it equal to zero. Then solve for x
*So for the derivative you get 4x^3-52x
4x^3-52x=0
4x(x^2-13)=0 *take out a 4x
4x=0 x^2=13 *divide
x=0, +/- sqrt of 13
^^So those are your critical numbers

Ex. 2) Locate the absolute extrema of the function f(x)=x^2+6x+4 on the closed interval [-6,6]
*Key words in this problem are "absolute extrema"..which tells you you're going to have to take the derivative of the function and set it equal to zero to find your x-values first. Then, you're going to plug your endpoints and the points of your interval into your original equation to find the max/mins.
*So for the derivative you're going to get 2x+6
2x+6=0
2x=-6
x=-3
*Plug in:
f(-3)=(-3)^2+6(-3)+4=-5
f(-6)=(-6)^2+6(-6)+4=4
f(6)=(6)^2+6(6)+4=76
**absolute max is (6,76); absolute min is (-3,-5)

Ex. 3) Determine whether Rolle's Theorem can be applied to f(x)=(x^2-16)/(x) on the closed interval [-16,16]. If Rolle's Theorem can be applied, find all values of c in the open interval (-16,16) such that f'(c)=0.
*First of all let's look at the function..Since they're asking us to use Rolle's Theorem we must first check if the function is BOTH continuous and differentiable before we can even use the theorem.
*The function is differentiable, butttttttttttt it is NOT continuous. It isn't continuous because there is a vertical asymptote at x=0. Therefore, the theorem CANNOT be applied.

Ex. 4) Determine whether the Mean Value Theorem can be applied to the function f(x)=x^2 on the closed interval [-8,2]. If the Mean Value Theorem can be applied, find all numbers c in the open interval (-8,2) such that f'(c)=f(2)-f(-8)/(2-(-8)).
*First check for differentiability and continuity.....and the function is both continuous and differentiable so you can continue using the theorem
*Next take the slope of the function and you should get this:
m=(4-64)/10
m=-6
*Now take the derivative of f(x)=x^2 and you should get 2x
*Now set your derivative equal to your slope (-6) and solve for x.
2x=-6
x=-3 *this is your c value

**Well for the most part I pretty much get the hang of all that we learned in Chapter 3. I could use some more practice though, but that's about it. Hope I do good on my test tomorrow!

BlogBlogBlogBlogBlogBlogBlogBlogBlogBlog

This week, we learned how to apply Rolle’s Theorem, the Mean Value Theorem, and the First Derivative Test to different problems. First of all, you need to know when to use all of them. Rolle’s Theorem tells you if there’s AT LEAST one max or min, but it doesn’t tell you how many or what they are while The Mean Value Theorem tells you that the slope of a secant line equals the slope of a tangent line between a and b and The First Derivative Test tells you were on the graph is there a max or min by telling you were the graph is decreasing or increasing. Using each of these different theorems is not hard at all. They just require practice and knowing when to use them so everything is memory of different forms of questions basically. Its not hard at all. Anyways, Thats all i have for today folks.

Week 2 Blog Prompt

What are the different ways a derivative can be interpreted as? So far what have you used the derivative for?

Blog #sihthcuottnac

This week, we learned about some theorems and whatnot

one of which was Rolle's Theorem

Rolle's Theorem states that for f(x), on the interval [a,b], if f(a) = f(b) then somewhere within that interval f'(c) = 0

unfortunately, that's all I've got...

Devin's Blog

Rolle’s Theorem states that “Let f be continuous on the closed interval [a,b] and differentiable on the open interval (a,b). If f(a) = f(b) then there is at least one number c in (a,b) such that f’(c) = 0.

Ex. Find the two x-intercepts of f(x) = x^2+5x+4

Set the equation equal to zero

F(x) = x^2+5x+4 = 0

Then factor the equation

F(x) = (x+4)(x+1) = 0

So, f(4) = f(1) = 0, and from Rolle’s Theorem you know that there exists at least one c in the interval (-1,-4) such that f’(c) = 0.

Take the derivative of the original equation

F(x) = x^2+5x+4 = 0

F’(x) = (2)x^2-1+5(1)x^1-1+4^0 = 0

F’(x) = 2x^1+5 = 0

And determine that f’(x) = 0 when x = -5/2, Note that this x-value lies in the open interval (-1,-5).

Rolle’s Theorem states that if f satisfies the conditions of the theorem, there must be at least one point between a and b at which the derivative is 0. There may of course be more than one such

Blog 9

This week we learned a couple of new theorems as we started the 2nd 9 weeks. yay.
We are on chapter 3, which involves extrema and critical numbers
Rolle's Theorem: Helps you find out if there is at least one max/min, but not how many

The Mean Value Theorem: Gives you the slope of a sec line that equals the slope of tan line between an a and b.

First Derivative Test: Shows where the max/min is by finding where it is increasing/decreasing

The first 2 theorems you need to make sure the function is continuous and differentiable

For the First Derivative Test, take the derivative and set = 0, then solve for x, find intervals, plug into the original equation.

Blog #9

This week, we learned how to apply 1) Rolle’s Theorem, 2) the Mean Value Theorem, and 3) the First Derivative Test to different problems. First of all, you need to know when to use all of them.
1) Rolle’s Theorem tells you if there’s AT LEAST one max or min, but it doesn’t tell you how many or what they are.
2) The Mean Value Theorem tells you that the slope of a secant line equals the slope of a tangent line between a and b.
3) The First Derivative Test tells you were on the graph is there a max or min by telling you were the graph is decreasing or increasing. It also finds the max or min.

Next, you need to know when to use them.
But, in order to use Rolle’s Theorem or the Mean Value Theorem, you first need to make sure your graph is continuous and differentiable.
1) If a problem asks “Find all values of c in the given interval such f’(c) = 0”, use Rolle’s Theorem.
*c = x
**Rolle’s Theorem generally uses x-intercepts.
2) If a problem asks “Find all values of c in the open interval [a, b] such that
f’(c) = f(b) – f(a) / b – a”, use the Mean Value Theorem.
***This theorem always uses the formula above.
3) If a problem asks “Find the relative extrema of…” or “Find the open intervals on which the graph is increasing or decreasing”, use the First Derivative Test.

Ex. 1) Let f(x) = x^4 – 2x^2. Find all values of c in the interval [-2, 2] such f’(c) = 0.
1. Is it continuous? Yes
2. Is it differentiable? Yes
The next step for Rolle’s Theorem is to check if the y-values are the same by plugging in the x-values from the interval into the original equation.
3. f(-2) = (-2)^4 – 2(-2)^2 = 8
f(2) = (2)^4 – 2(2)^2 = 8
Are the y-values the same? Yes
Next, take the derivative and set it equal to 0, then solve for x.
4. f’(x) = 4x^3 – 4x
4x^3 – 4x = 0
4x(x^2 – 1) = 0
x = 0, 1, -1
Therefore, c = 0, 1, -1
Finally, justify your answer.
By Rolle’s theorem, the function is continuous, differentiable, and f(-2) = f(2), therefore there is at least one max or min on [-2, 2].

Ex. 2) Given f(x) = 5 – (4/x), find all values of c in the open interval [1, 4] such that f’(c) = f(4) – f(1) / 4 – 1.
1. Is it continuous? Yes
2. Is it differentiable? Yes
The next step of the Mean Value Theorem is to find the slope by plugging into the formula.
3. 5 – (4/4) = 4
5 – (4/1) = 1
4 – 1 = 3
4 – 1 / 3 = 1
Next, take the derivative and set it equal to the slope you just found, then solve for x.
4. f’(x) = 4/x^2
4/x^2 = 1
x = 2, -2
Therefore, c = 2, -2
By the Mean Value Theorem, the function is continuous and differentiable on the interval, therefore there is some value (c = 2) where the derivative equals the slope between 1 and 4.

Bloggy mc Blog Blog

So of course we are starting a new section and its kind of graph related you could say, we started off by learning what critical numbers were: max's and mins. So if the question may ask to find a critical number, you would be looking for the max or min. Now there are two theorems that you use. The two theorems are Rolle's theorem and Mean Value theorem. For each of these theorems your going to have to figure out if it's continuous and differentiable before continuing. Rolle's theorem deals with intervals, and it's to tell the max and min. Mean value theorem is dealing with slope.. I'll explain section 3.3 since I know it. 3.3 is dealing with the first derivative test. The first derivative test has 3 steps to it: 1. Take derivative=0. 2. Solve for x. 3. Set up intervals, then plug back into derivative. (This will tell you whether the number is increasing or decreasing.) and i don't really have any of my math stuff with me so i cant really give you any examples :/

blog #suluclacdne

this week we learned about rolle's theorem. rolle's theorem states that for f(x) on the interval [a,b] if f(a) = f(b) then somewhere within that interval f '(c) = 0.

for example: on the interval [1,-1] f(x) = 3x^4 + 2x^2

f(1) = 3 and f(-1) = 3

this means that 6x = 0.

x = 0.

there is a minimum at zero.

Alaina's blog, 24 Oct 2010

This week we talked about several topics like Rolle’s Theorem, the Mean Value Theorem, and the First Derivative Test. The first topic we learned was Rolle’s Theorem. This tells you whether or not you have at least one max or min on the interval.
-let f be continuous on the closed interval [a,b] and differentiable on the open interval (a,b).
If f(a)=f(b)
then there is at least one number c in (a,b) such that fprime(c)=0.
*you must test for continuity and differentiability before using Rolle’s Theorem. You must check if the y-values match. (plug in and check).
**typically Rolle’s Theorem uses x-values [a,b] to find y-values.
***fprime(x)=0 implies max or min

Ex 1: find the 2 x-intercepts of f(x)=xsquared -3x +2 and show that fprime(x)=0 at some point between the 2 x-intercepts.
1. continuous? Yes
2. differentiable? Yes
3. xsquared -3x +2=0
(x-2)(x-1) same y-values
x=1,2 [1,2]
By Rolle’s theorem, the function is continuous, differentiable, and f(1)=f(2). Thus there must be at least one max or min (fprime(x)=0) on the interval [1,2].

10/24/10

Soooo, this week the first thing we learned was Rolle's Theorem-which tells you IF there is a max or min on an interval. To apply this theorem you must first check the function for continuity and differentiability, and the y-values have to match. *Also, Rolle's theorem tells you that you have AT LEAST 1 max or min, it does NOT tell you how many you have. Next, we went over the Mean Value Theorem, which says that on an interval [a,b] there must be some x-value where the derivative equals the slope between a and b. To apply this theorem, you must also check for continuity and differentiability. Last, we learned how to apply the First Derivative Test, which is used to actually find the max/mins and what interval they are on. **Anddddddd, I'm not going to lie, all of these theorems aren't extrememly difficult, but they're giving me a hard time especially with the more complicated problems. So, yeah I'm going to need some help before the test. Anywayyyyyyy, here are some examples:

Ex. 1) Determine whether Rolle's Theorem can be applied to f on the closed interval [a,b]. If it can be applied, then find all values of c in the open interval (a,b) such that f'(c)=0.
f(x)=(x^2-2x-3)/(x+2) [-1,3]
*First, check for continuity and differentiability of the function.
*Now what you probably notice is that x=-2 (the denominator of the fraction) is a vertical asymptote, making the function NOT continuous....HOWEVER, -2 is not in the interval [-1,3] so it does not matter. Rolle's Theorem can still be applied...Oh and the function is differentiable too
*Now, check to make sure your y-values are the same..So to do that, you just plug -1 into the original equation and plug 3 into the original equation and make sure you get the same # answer.
f(-1)=0
f(3)=0
*Now you take the derivative of the function and you should get:
(x+2)(2x-2)-[(x^2-2x-3)(1)]/(x+2)^2
=(x^2+4x-1)/(x+2)^2
*Then you set the top^of the fraction equal to zero to find you're x-values.
x^2+4x-1=0
This doesn't factor, so you'll have to use the quadratic formula and you'll end up getting this:
= -2+/-sqrt of 5
**Your c value is only -2+sqrt of 5 because -2-sqrt of 5 is not on the interval
***Soooo, by Rolle's Theorem, the function is continuous, differentiable, and f(-1)=f(3); Therefore at c=-2+sqrt of 5, there must be either a max or a min

Ex. 2) Determine whether the Mean Value Theorem can be applied to f on the closed interval [a,b]. If it can be applied, find all values of c in the open interval (a,b)..if not, explain why not
f(x)=x^3+2x [-1,1]
*Firsttttttttttt, check for differentiability and continuity....and the function IS in fact differentiable and continuous
*Next, you must find the slope..To do that you use f(b)-f(a)/b-a..so you should get:
f(1)-f(-1)/1+1
=3+3/2
*So your slope is 3
*Now you take the derivative of your original equation and you should get 3x^2+2
*Nowwwwww, you take the derivative you just got and set it equal to your slope and solve for x like this:
3x^2+2=3
x=+/-sqrt of 3/3
Soooo that's^^^^your c values

**As for what I didn't understand this week..which was quite a few things. I need help with the problems where you have to just look at the graph and determine everything and I need help with problems on the study guide like #13, #16, #19, #20...and many others. Put it this way, I need a lot more practice with this because half the time I have no clue what I'm doing

Blog #9

This week we started fresh with the 2nd nine weeks. 25% done school! We started and continued learning material from Chapter 3. At the beginning of the week we learned Rolle's Theorem and the Mean Value Theorem. I am going to explain how to use these the Rolle's Theorem.

Rolle's Theorem

Let f be continuous on the closed interval [a,b] and differentiable on the open interval (a,b).

If f(a) = f(b), then there is at least one number c in (a,b) such that f ' (c) = 0

*Test for continuity and differentiability before using Rolle's Theorem.

Rolle's Theorem tells if there is a max or a min on an interval, and typically, Rolle's Theorem uses x-intercepts.

Example:

Let f(x) = x^4 - 2x^2. Find all values of c in the interval (-2, 2) such f ' (c) = 0.

Step 1: Is the function continuous? Yes, because it is a polynomial.

Step 2: Is the function differentiable? Yes, because there are no points of non-differentiability or items in the function that make it not differentiable.

Step 3: Plug in the x and y from the open interval into the original equation and see if they both have the same output.

f(-2) = (-2)^4 - 2(-2)^2 = 8

f(2) = (2)^4 - 2(2)^2 = 8

Now take the derivative of the original function and set it equal to zero. Then find the x-intercepts.

4x^3 - 4x = 0

4x(x^2 - 1) = 0

x = 0, -1, 1

By Rolle's Theorem, the function is continuous, differentiable, and f(-2) = f(2), then there must be at least one max or min on [-2, 2] where (f ' (x) = 0).

Blog 9.

Alright, so this week was the start of the second nine weeks, and we started chapter three. Which doesn't seem like it's going to be too hard, although some of the rules are hard to remember. Anyways, we started off by learning what critical numbers were, which are max's and mins. So when something ask to find a critical number, you would be looking for the max or min. Now there are two theorems that you use, I don't have my binder with me so I'm not gonna go into detail to explain them because I don't want to give the wrong information. The two theorems are Rolle's theorem and Mean Value theorem. For each of these theorems your going to have to figure out if it's continuous and differentiable before continuing. Rolle's theorem deals with intervals, and it's to tell the max and min. Mean value theorem is dealing with slope. That's pretty much all I have to say about that since I don't have my binder. I'll explain section 3.3 since I know it. 3.3 is dealing with the first derivative test. The first derivative test has 3 steps to it: 1. Take derivative=0. 2. Solve for x. 3. Set up intervals, then plug back into derivative. (This will tell you whether the number is increasing or decreasing.)

Find the open intervals on which f(x)=x^3-3/2^x^2
1. 3x^2-3x=0
2. 3x(x-1)
x=0,1
3. (-infinity, 0) u (0,1) u(1,infinity)
4. f(-1)=+ f(1/2)=- f(2)=+
**Increasing (-infinity,0) u (1,infinity)
**Decreasing (0,1)
Now for something I don’t understand, I am still very confused with trig functions when it comes to problems like these.

Taylor Blog #9

This week we covered a few major lessons. These lessons included Rolle’s Theorem, the Mean Value theorem and the first derivative test. So I suppose I’ll briefly describe how to work with each and Ill provide an example problem for each.
Rolle’s Thorem:
Rolle’s Thorem is used to discover whether there is a max or a min on the given interval.
Steps:
Check differentiability
Check for continuity
If the problem is not differentiable or if the problem is not continuous Rolle’s thorem cannot be applied.
If the problem is differentiable and is continuous then you will continue to work the problem by:
Test that the y values match by plugging interval numbers into the original problem
Take the derivative and set equal to zero
Solve for x
**Remember f’(c)= 0 must be read as max or min and therefore you must use rolle’s thorem
Example:
Let F(x)=x^4-2x^2 find all values of c in the interval [-2,2] such that f’(c)=0
Check differentiability
Differentiable? Yes.
Check for continuity
Continuous? Yes
Test that the y values match by plugging interval numbers into the original problem
F(-2)= -2^4-2(-2)^2
16-2(4)
16-8
=8
F(2)= 2^4-2(2)^2
16-2(4)
16-8
=8
Therefore, yes
Take the derivative and set equal to zero
4x^3-4x=0

Solve for x
4x(x^2-1)=0
X=0 X=-1 X=1
Therefore by Rolle’s thorem c= 0, -1, 1
By Rolle’s theorm the function is continuous, differentiable, and f(1)=f(2) thus there must be one max or min on the interval [-2, 2] at f’(x)=0

Mean Value Thorem: The mean value theorem is used to prove that on the interval there must be some x value where the derivative equals the slope between the points of the interval
Steps:
Check Differentiability
Check Continuity
If the problem is not differentiable or if the problem is not continuous the Mean Value theorem cannot be applied.
If the problem is differentiable and is continuous then you will continue to work the problem by:
Check Slope (((( f’ (C)= f(b)-f(a)/b-a))))
Take the derivative and set equal to zero
Solve for x
Example: Given f(x)= 5-(4/x) find all values of c in the open interval (1,4) such that f’(c)= f(4)-f(1)/ 4-1
Check Differentiability
Differentiable? Yes
Check Continuity
Continuous? Yes
Check Slope (((( f’ (C)= f(b)-f(a)/b-a))))
4-1/4-1 = 1

Take the derivative and set equal to slope
-4(-1)x^-2
4x^-2
4/x^2= 1
Solve for x
X=+/- 2
Therefore by the mean value theorem, the function is continuous and differentiable on the interval therefore there is a value c=2 where the derivative = slope on the interval (1,4)
The first derivative test:
Used to determine whether it is specifically a max or a min on the interval
KEY WORDS TO RECOGNIZE WHEN TO USE THE FIRST DERIVATIVE TEST
Increasing, decreasing, max, min, horizontal tangent, change direction
Steps:
Take derivative and set equal to zero
Solve for x
Set up intervals (-infinity, pt)U(pt, infinity)
Plug in value on the interval into the derivative ((if you get a positive number the function is increasing, If you get a negative number the function is decreasing)) (( A max is increasing then decreasing, A min is decreasing then increasing))
Example:
Find the open intervals on which f(x)= x^3-3/2x^2 is increasing or decreasing
Take derivative and set equal to zero
3x^2-3x=0
Solve for x
3x(x-1)=0
X=0,1
Set up intervals (-infinity, pt)U(pt, infinity)
(negative infinity,0)U(0,1)U(1,infinity)
Plug in value on the interval into the derivative
F’(-1)= 3(-1)^2-3(-1)=6
Positive therefore increasing
F’(1/2)= 3(1/2)^2-3(1/2)= -3/4
Negative therefore decreasing
F’(2)= 3(2)^2-3(2)= 6
Positive therefore increasing
Therefore increasing: (negative infinity, 0) U (1, infinity)
Decreasing: (0,1)
Max: at 0
Min: at 1

Week 8 Blog Prompt

What are the steps to a related rate problem? What are the key words you look for and what do they mean? Give an example of a related rate problem and solve it.

Blog for 10/17

well, this week we learned how to find extremas

Definition of Extrema
Let f be defined on an interval I containing c. f(c) is the minimum of f on I if f(c) < f(c) for all x in I
f(c) is the minimum of f on I if f(c) > f(c) for all x in I
The minimum and max of a function on an interval are the extreme values or extrema of the function on the interval. The minimum and max of a function on an interval are also called the absolute min and absolute max on the interval, respectively.

Definition of Relative Extrema
If there is an open interval containing c on which f(c) is a maximum, then f(c) is called a relative maximum of f.
If there is an open interval containing c on which f(c) is a minimum, then f(c) is called a relative minimum of f.

Definition of Critical Number
Let f be defined at c. If f '(c) = 0 or if f ' is undefined at c, then c is a critical number of f.

Example:

Find the extrema of f(x) = 3x3 - 4x4 on the interval [-1, 2].

Find the critical numbers
f(x) = 3x3 - 4x4   {Problem}
f '(x) = 9x2 - 16x3   {Differentiate}
9x2 - 16x3 = 0   {Set f '(x) = 0}
x(9x - 16) = 0   {Factor}
x = 0, 16/9   {solve to get the critical numbers}

Blog #8

This week in Calculus, we learned how to find extreme values, critical numbers, absolute max and min, and relative max and min.

First, we need to remember the definitions of all of these terms:
Extreme values – max and min
Absolute max and min – the highest and lowest point on an interval
Relative max and min – any max or min not on an interval
Critical number – any max or min (often referred to as x = c)

Now, to find a critical number, etc. is the same as finding a horizontal tangent (something we did in the last chapter). You take the derivative, then set equal to zero. If you have an interval and it asks for the max and min, plug your x-values you get from solving the derivative and your endpoints from the interval into the original equation. The biggest number is your max and the smallest number is your min.

Ex. Find the max and min on the graph f(x) = cos π x on the interval of [0, 1/6].
f ’(x) = (-sin π x) (π) = -π sin π x
-π sin π x = 0
sin π x = 0
π x = sin^-1(0)
π x = 0, π, 2π
x = 0, 1, 2
f(0) = cos π (0) = cos 0 = 1
f(1/6) = cos π (1/6) = cos π/6 = ½
f(1) = cos π (1) = cos π = -1
f(2) = cos π (2) = cos 2π = 1
max = 1
min = -1

Justin 8

This week we wrapped up chapter 2 and went into chapter 3

what we learned so far is sort of a repeat of ch 2, finding extrema

THE RULES:
1. Find the critical numbers
2.Evaluate at each critical number
3.Evaluate at each endpoint
4.The least of the values is the min. the greatest is the max
5.Anything not differentiable is a critical number

we work out problems in this chapter by looking at graphs (sometimes)

the absolute max is the highest point on the interval
the absolute min is the opposite (duh)
the relative max is any max that is not on the interval
the relative min is any min '' '' ''
a critical number is any max or min typically referred to as x=c

Find the extrema of f(x)=3x^4 - 4x^3 on the interval [-1,2]

12x^3 - 12x^2 = 0
12x^2 (x-1)=0
x=0,1

To determine max/min on an interval
f(-1)=7
f(0)=0
f(1)=-1 absolute min
f(2)=16 absolute max

Blog #topsgolbredrum

okay time to spout some premium quality bs yet again. this week we learned about critical numbers and what they are used for. they are used to find the maximum and minimum of a graph. and for he misled purpose of writing toward a word count i will explain how. in order to find your critical numbers, you must take the derivative of a function and set it equal to zero. critical numbers are most often referred to using the expression x = c. critical numbers also represent the horizontal tangents of an equation.

for example: 2x^2 - 4x

4x - 4 = 0

4x = 4

x = 1

there is a horizontal tangent at one... congratulations

Alaina's blog, 17 October 2010

This week we had exams, and starting on Thursday, we began notes on finding critical numbers, extreme values, maximums and minimums, and horizontal tangents all by taking the derivative, setting it equal to zero and solving for x.

Guidelines for finding extrema on a closed interval:
-to find the extrema of a continuous function f on a closed interval [a,b] use the following steps:
1. find the critical numbers of f in (a,b).
2. evaluate f at each critical number in (a,b).
3. evaluate f at each endpoint in (a,b).
4. the least of these values is the minimum and the greatest is the maximum.
5. anywhere the function is not differentiable is a critical number.

Key words:
Extreme values- maximums and minimums
Absolute max- highest point on an interval
Absolute min- lowest point on an interval
Relative max- any max not on the interval
Relative min- any min not on the interval
Critical numbers- any max or min typically referred to as x=c

**always check endpoints on an interval!

Ex: f(x) =2x-3x^ (2/3) on [-1, 3]
F prime(x) = 2-2x^ (-1/3) =0
2-2x^ (-1/3) =0
-2x^ (-1/3) =-2
(x^ (-1/3))^3 =-2/-2=1^ 3
x=1

f(-1)= 2(-1) -3(-1)^(2/3)= -5-> min
f(1)= 2(1)-3(1)^(2/3)= -1
f(3)= 2(3)-3(3)^(2/3)= -.24
f(0)= 2(0)-3(0)^(2/3)= 0-> max

Blog #8

This week wasn't too bad at all, it passed by kinda really fast for me. Anyway, we started Chapter 3, finally, and we started off learning the guidelines for finding extrema on a closed interval.

To find the extrema of a continuous function f on a closed interval [a,b], use the following steps.

1. Find the critical numbers of f in (a,b).

2. Evaluate f at each critical number in (a,b).

3. Evaluate f at each endpoint of [a,b].

4. The least of these values is the minimum. The greatest is the maximum.

There are different terms in 3.1 that we should learn.

Extreme values - mins and maxes

absolute max - is the highest point on an interval

absolute min - in the lowest point on an interval

relative max (local) - any max not on an interval

relative min (local) - any min not on an interval

critical numbers - any max. or min. typically referred to as x=c

To find a critical number, extreme value, max, min, or horizontal tangent, take derivative and set it equal to 0.

Example:

Find the critical values of the function

f(x) = 4x^3 - 36x

12x^2 - 36

12x^2 - 36 = 0

12x^2 = 36

x^2 = 3

x = +/- sqrt 3 <------FINAL ANSWER

It's that simple.

Now, I'm not too sure about how to do extrema with trig functions and stuff. It's a little complicated when you have to use formulas, identities, trig functions, and other stuff all at the same time.

Taylor Rodriguez blog #8

This week we started chapter three. We began learning about extreme values, absolute max, absolute min, relative max, relative min, and critical numbers.
Extreme Values are defined as maximums and minimums
Absolute Max is defined as the highest point on an interval
** notice that absolute max is the highest point on a given interval not the highest point on the graph
Absolute min is defined as the lowest point on an interval
**notice that absolute min is the lowest point of a given interval not the lowest point on the graph
Relative Max (also referred to as local max) is defined as any max NOT on an interval
Relative Min (also referred to a local min) is defined as any min NOT on an interval
Critical Numbers are defined as any max or min typically referred to as x=c
When trying to solve problems concerning a critical number, extreme value, max, min, or horizontal tangents take the derivative and set equal to zero.

Example #1:
Find the critical values of f(x)= 9(x^2-3)/x^3
9x^2-27/x^3
Take the derivative
X^3(18x)-[(9x^2-27)(3x^2)]/ x^6
18x^4-[27x^4-81x^2]/x^6
18x^4-27x^4+81x^2/x^6
-9x^4+81x^2/ x^6
9x^2(-x^2+9)/X^6
9(-x^2+9)/x^4
Set equal to zero and solve
((when setting a fraction equal to zero you are really only setting what is in the numerator equal to zero))
9(-x^2+9)=0
-9x^2+81=0
-9x^2=-81
X^2=9
X=+/- 3

Example #2:
Find the extrema of f(x)= 3x^4-4x^3 on the interval [-1,2]
4(3)x^3-3(4)x^2
12x^3-12x^2=0
12x^2(x-1)=0
X=0,1

To determine max or min on an interval plug into original and the biggest result will be the max
and the smallest result will be the min
f(-1)= 3(-1)^4-4(-1)^3=7
f(0)=3(0)^4-4(0)^3=0
f(1)= 3(1)^4 – 4(1)^3= -1 ((smallest so absolute min))
f(2)= 3(2)^4-4(2)^3= 16 ((largest so absolute max))

10/17/10

This week we had exams and then on Thursday we started Chapter 3. We learned how to find critical numbers, extreme values, max and mins, and horizontal tangents..simply by taking the derivative and setting it equal to zero! Also, to determine the max or min on an interval, you plug in the numbers of your interval and the numbers of your endpoint into the original equation. Overall, the concepts this week were pretty easy..I just need to practice with it some more. So here are some examples:

Ex. 1) Find any critical numbers of the function f(x) = 4x / x^2+1
*Since they're asking you to find critical numbers, the first thing you have to do is take the derivative of the function
*So, using the quotient rule, you get:
*(x^2+1)(4) - [(4x)(2x)] / (x^2+1)^2
Simplifying that you get:
-4x^2+4 / (x^2+1)^2
*To find the critical numbers, you take the numerator of the fraction and set it equal to zero.
So you get:
-4x^2 + 4 = 0
x^2 = 1
*So your critical numbers are x = 1, -1

Ex. 2) Locate the absolute extrema of the function on the closed interval for f(x) = (2x+5)/3 on the interval [0,5]
*The first thing you want to do for this problem is find your extreme values (which is the same as finding critical numbers) Sooooo, take the derivative of the function and set it equal to zero
*So for the derivative you get:
(3)(2) - [(2x+5)(0)] / 9
=6/9
f'(x) = 2/3 (*Since your derivative is a number and not a function, you have NO critical numbers)
*So you just need to plug in the numbers of your interval into the original equation to find the max and min
*So you get:
f(0) = 2(0)+5 / 3 = 5/3
f(5) = 2(5)+5 / 3 = 5
*So your answers are:
(0,5/3) min
(5,5) max

Ex. 3) Locate the absolute extrema of the function on the closed interval for f(x) = x^3-3/2x^2
on the interval [-1,2]
*First thing you have to do is take the derivative and set it equal to zero to find your critical numbers/extreme values
*So you get:
f'(x) = 3x^2 - 3x
*3x^2 - 3x = 0
3x(x-1) = 0
x = 0, 1 (extreme values)
*Now you plug in the numbers in the interval and the numbers of your endpoint into the original equation
*f(-1) = -1-3/2(-1)^2 = -5/2
f(2) = 8 - 3/2(2)^2 = 2
f(0) = 0
f(1) = 1-3/2 = -1/2
*So your answers are:
(2,2) max
(-1,-5/2) min

**So pretty much everything we learned this week was easy for me. The only thing I don't understand is how to find critical numbers and everything else when you are just given a graph to look at in the book...Alsoooo, I need to refresh my memory on Chapter 8 from Advanced math, hahah

Blog 8

Soo, I did pretty well on this nine weeks exam :) hhaa, anyways. We started learning chapter three this week and it seems like its going to be pretty tough because it goes back to everything we learned so far in chapter’s one and two, and even stuff that we learned in advance math. Right now we are learning how to find extreme values. Extreme values are the max and mins of a function, or graph. The absolute max is the highest point on an interval. The absolute min is the lowest point on an interval. We also learned how to find relative maxs and mins, which are any max or min not on an interval. If it asks you to find a critical number, you’re looking for any max or min. **Keep in mind anywhere the function is NOT differentiable it is a critical value. **To find a critical number, extreme value, max, min, or horizontal tangent, take derivative and set equal to 0. **Also when dealing with a fraction you would take the derivative by doing the quotient rule, and then set the top equal to 0. The bottom doesn’t really matter in a fraction. **If it gives you an interval you will take the derivative of the function, set it equal to 0, then check to see if the numbers are in the interval. Then you will plug the numbers you found, and the interval numbers back into the original function. The highest point will be your abs. max, the lowest your abs. min.

Ex: Find the extrema of f(x)=3x^4-4x^3 on the interval [-1,2]
12x^3-12x//62=0
12x^2(x-1) x=0, x=1

*Now you’re going to plug back into the original equation.
F(-1)=3(-1)^4-4(-1)^2=7
F(0)=3(0)^4-4(0)^3=0
F(1)=3(1)^4-4(1)^3=-1
F(2)=3(2)^4-4(2)^3=16
**-1 Abs. min, 16 abs. max

Now for something I don’t really understand. I don’t understand the trig functions ones where you have to use the quadrants then convert to radians. I know how to use the quadrant and convert to radians, I just don’t know how to get there.

Week 7 Blog Prompt

What concept do you feel like you mastered this nine weeks? What concept did you struggle with the most? Why? What can you change this nine weeks in your study habits, etc to improve your grade?

Devin's Blog

Related rates

X and y are both differentiable and they are related by an equation.

Use the Chain rule to find the derivative of each side of the equation.

Make sure to differentiate the equation in respect to t.

Area Derivatives

First find the correct area equation for the problem.

Then take the derivative of both sides of the equation.

Make sure that you differentiate with respect to t.

Then Chain rule the rest of the problem.

Then substitute and solve for the variable.

Volume Derivatives

First find the correct volume equation for the problem

Then take the derivative of both sides of the equation

Make sure that you differentiate with respect to t

Then substitute in your given variables

And solve for your unknown

Speed Problems

Most of these problems are triangularly related

Use the Pythagorean theorem to get your equation

The take the derivative of the equation

Be sure to take the derivative with respect to t

Then solve for your unknown

Then substitute in your variables

Finally simplify your answer

Ex. Y=x^2+9 x=7 dx/dt=2

Dy/dt= d/dt(x^2+9)

Dy/dt=2xdx/dt

Dy/dt= 2(7)(2)

Dy/dt=28

alaina's blog, 10 oct. 2010

Alright, this week is exam week and I’m nervous about the exam for this class. Actually, I'm FREAKING OUT. I’m hoping to make a good grade, but I guess we will have to see. Anyway…

I’ll explain finding tangent lines when it gives you an equation or formula, and a point. When this happens you’re going to do what you would usually do which is figure out what kind of problem you’re dealing with. Once that’s done you’re going to set up your problem and take the derivative, then solve. After getting an answer, which will be what dy/dx is equal to, you’re going to plug in the point they give you. Then you will have to plug into point slope form.

Here’s an example:Find the tangent line to the graph given by x^2(x^2+y^2) at the point (square root of 2/2, square root of 2/2).
*First thing you do is distributeX^4+x^2y^2=y^2
**Notice it is now product rule, so now you can set up & take the derivative4x^3+x^2(2ydy/dx)+y^2(2x)=2ydy/dx
*Solve and set dy/dx on one sideDy/dx=4x^3+2xy^2/2y-2x^2y
*Now plug inDy/dx=4(square root 2/2)^3+2(square root 2/2)(square root 2/2)^2/2(square root 2/2)-2(square root 2/2)^2(square root 2/2)*Solve to get slopeM=3
*Put in point slope
ANSWER: y-square root of 2/2=3(x-square root of 2/2)

One of the things that I’m still not too comfortable with is the word problems. When it doesn’t tell me what the problem is, or what formula to use, I get confused and can’t figure out how to work it. I can set it up, but I get a little confused from there. I guess I just need more practice; it also will help me once I get all the formulas.

Blog # blah

I'mma be straight up with this, I don't know math, I'm gonna fail the exam tomorrow, I can't remember anything we did last week, and my Aunt's about to DIE, this is f&#**&% insane and I just want this week to end so things can go back to being semi-normal
:(

Blog #kanweestopdointhisyet

okay this week was mostly review with the assistance of a few super complex word problems. we pretty much just went over all of the rules that we have been learning this nine weeks and worked on our skills in applying them to harder problems.

we had a few problems in which we had to apply the rule of constants which is pretty easy. all it says is that all constants have derivative of zero.

we also did a little bit of work using the power rule which states that when you have a variable you take the exponent and multiply it by whatever is in front of the variable. then, you subtract one from the exponent and leave it there.

for example: x^4 becomes 4x^3. also pretty easy.

we also had a few problems in which we had to use quotient rule which is just a complicated way of deriving a fraction. what you do is you take the derivative of the top and multiply it by the bottom and then you subtract from that the derivative of the bottom multiplied by the top and thats all over the bottom squared.

Blog 7

So this week we sorta reviewed chapter 2, derivatives, and tried out related rate problems, which are hard to grasp depending on the structure of the word problems.

Finding a derivative of a function, in my words, is like this: the more that is going on in the problem, the more work needed and more steps taken to find it. Ex: the chain rule

if you're asked to find the slope of a tangent line, dy/dx, f'(x), instantaneous velocity, it simply means take the derivative.

My interpretation on the Rules of derivatives:
1. if there is a constant, the derivative of a constant=0 f(5)=5 f'(5)=0

2. when there is an exponent on a variable, you multiply by the exponent and subtract 1 from the exponent Ex: f'(3x^2)=6x

3.when there is a fraction, bottom(f'(top)-[top(f'bottom)]/bottom^2

4.when there is multiplication involved, 1st(f'2nd)+2nd(f'1st)

5.when there is more than one operation (3x+4)^2, bring exponent to the front, recopy inside, multiply by f'inside
2(3x+4)(3)
6(3x+4)=18x+24

6.when trig is involved, remember those identities to simplify as much as possible (tricky)

7.when there is more than one variable (implicits) you solve for dy/dx using algebra

8.when there is a word probelm, *sigh* related rates can get confusing depending on the setup and the geometry used (like the ones about the pool...yuck), you start by finding the rate, setting up the information, then start using geometry with derivatives and the complex and the formulas and the oh you know what i mean!
well exams are tomorrow so back to studying

Blog #?-Dustin

Well, this past week was review week, so nothing new was covered. I'm pretty sure everyone in the class was pretty stressed over the insane word problem homework, but anyways on to some review stuffz.

Some of the beginning rules of derivatives were the constant rule and the power rule. Those are pretty simple:

Constant Rule: It states that the derivative of any constant is 0. Easy enough, right? WRONG!

Power Rule: This one is also quite simple, the derivative of any number/variable while its by itself is figured out by using the following steps. Take the exponent of the variable/number and bring it to the front and subtract one from the original exponent to form the new one. An example would be:
Ex. Find the derivative of x^y.
First you move y to the front. Then subtract one from y to form the new exponent and you get...
(xy)^y-1

Well those are the simplest ones. Then you have rules like the quotient rule and product rule. Those are also pretty simple.

Quotient rule: When you have a fraction to derive. Take the( bottom(top derivative)-top(bottom derivative))/bottom^2

Product rule: When you derive one thing times another. Take the derivativeoffirst(second)+first(derivativeofsecond)

Those are also pretty easy. You also have many trig rules that I don't even feel like typing. Then theres the chain rule which is like the power rule except you have an equation and you multiply by the derivative of the inside afterwards. Well, I guess thats about it for a simple derivative review. Theres also implicits, rate of changes, other nonsense, and limits on the exam, but you already know that. Now, to go about my regular test ritual and study for it by doing nothing=]

Blog #7

Ok this week was pretty much a review of derivatives, but we also learned how to solve related rate problems. Well, truthfully, I didn't learn how to do the related rate stuff, I just kinda went with the flow of class, I don't think anybody really understood how to do those problems this week, and we have an exam tomorrow. Ooh this is gonna be fun :/ Anyway, I'll review some "simple" derivative work.

Find the derivative of sin y. Find the largest interval about the origin for which y is a differentiable function of x.

sin y = x

cos y (dy/dx) = 1

dy/dx = (1)/(cos y)

pi/2 is less than y is less than 3 pi/2

Determine the slope of the tangent line to the graph of x^2 + 4y^2 = 4 at (sqrt(2), -1/sqrt(2))

2x + 8y(dy/dx) = 0

8y(dy/dx) = -2x

dy/dx = -2x/8y

dy/dx = -2(sqrt(2)/8((-1)/(sqrt(2))) = 2(2)/8 = 4/8 = 1/2

slope = 1/2

Like I said on the top of the blog, I don't understand the rate of change problems. And with this exam coming up TOMORROW, i don't know what's gonna happen, but oh well.

Blog 7?

Alright this week is exam week:\ and I’m nervous about the exam for this class. I’m hoping to make a good grade, but I guess we will have to see. Anywaysssss… I’ll explain finding tangent lines when it gives you a problem, and a point. When this happens you’re going to do what you would usually do which is figure out what kind of problem you’re dealing with. Once that’s done you’re going to set up your problem and take the derivative, then solve. After getting an answer (which will be what dy/dx is equal to) you’re going to plug in the point they give you. Then you will have to plug into point slope form.

Here’s an example:
Find the tangent line to the graph given by x^2(x^2+y^2) at the point (square root of 2/2, square root of 2/2).
*First thing you do is distribute
X^4+x^2y^2=y^2 **Notice it is now product rule, so now you can set up & take the derivative
4x^3+x^2(2ydy/dx)+y^2(2x)=2ydy/dx
*Solve and set dy/dx on one side
Dy/dx=4x^3+2xy^2/2y-2x^2y
*Now plug in
Dy/dx=4(square root 2/2)^3+2(square root 2/2)(square root 2/2)^2/2(square root 2/2)-2(square root 2/2)^2(square root 2/2)
*Solve to get slope
M=3
*Put in point slope
ANSWER: y-square root of 2/2=3(x-square root of 2/2)

One of the things that I’m still not too comfortable with is the word problems. When it doesn’t tell me what the problem is I get confused and can’t figure out how to work it. I can set it up, but I get a little confused from there. I guess I just need more practice; it also will help me once I get all the formulas. From what I understand the only word problems that will be on the exam are the triangle ones, volume, and Area.

10/10/10

Well this week we worked with related rate word problems, which I must say are not fun at all. They're called related rate because somewhere in that problem you're going to have to take a derivative. There's a lot to decipher within one word problem..First you have to figure out what type of information they're giving you and what formula you're going to have to use. Then you have to figure out what exactly you're looking for (what your answer would be) based on that information.
Here are some key words in word problems:
*"a rate of 4.5 cubic feet per minute" = cubic feet indicates VOLUME
*"how fast is something rising" = rising indicates height, which means you'd be looking for dh/dt
*"radius of" = r
*"s" = hypotenuse of a triangle
*"depth" = height
*"altitude" = height

Here's an example:
Volume: All edges of a cube are expanding at a rate of 6 centimeters per second. How fast is the volume changing when each edge is a.) 2 centimeters and b.) 10 centimeters?
*Okay first let's identify what information we're given.
They tell us that it's expanding at a rate of 6 centimeters per second, so that means:
dr/dt = 6 cm/sec
Then you notice that they're looking for the volume of a cube, so you're going to need that formula which is V = x^3 (Well in this case it would be V = r^3..but it's the same thing)
Lastly they ask for how fast the volume is changing..."how fast" is the same thing as rate, and they're talking about volume so that means that they want you to find dv/dt
*So the first thing you want to do when solving this problem is take the derivative of your formula V = x^3
*So you get dv/dt = 3x^2 dx/dt
*Now part a.) wants you to find dv/dt when the edge is 2 centimeters..So all you would do is plug in 2 for x and plug in 6 for dx/dt
And you would get dv/dt = 3(2)^2(6) = 72 cm^3/sec
*Remember, it's cm^3 because volume is always cubed
*Part b.) wants dv/dt when the edge is 10 centimeters...So for this all you do is plug in 10 in place of where you plugged in 2 for part a.
So you get dv/dt = 3(10)^2(6) = 1800 cm^3/sec

And there you go

Blog #7

This week, we learned about word problems that have to do with rate of change (aka derivatives). There are steps that you can take to decipher the word problems and make the easier to understand. First, you need to identify all of the information the word problem gives, such as variables and what you’re actually trying to find. Once you do that, then you need to find a formula that connects all of your variables and your shape together (such as the area of a triangle or the volume of a cube). After this, solve for your desired rate with respect to t (such as dA/dt for the rate of area changing or dV/dt for the rate of volume changing). And lastly, plug in all of your variables to get your answer.
*Don’t forget to include your units (such as cm/min or ft^3/sec).

Ex. The radius r of a circle is increasing at a rate of 4 centimeters per minute. Find the rates of change of the area when r = 8 cm.

First of all, we know the following info: r = 8 & dr/dt = 4 cm/min
We also know we are looking for: dA/dt = ?
Second, because we are looking for the rate of change of the AREA, we know where using the area of a circle formula: A = π(r^2)
Now, solve for dA/dt by taking the derivative of the area formula: dA/dt = 2π r dr/dt
**Remember to put dr/dt after you take the derivative of r because it is an implicit derivative.
Finally, plug in your variables: dA/dt = 2π (8)(4) = 64π cm^2/min
***cm is squared because area is always squared, just like volume is always cubed.

blog #7

This week we learned about solving related rate word problems.
There are a set of steps to follow when solving a related rate problem but first you need to know how to recognize a related rate problem.
A related rate problem is one that involves something per something.
I.E. Gallons per minute
Cubic feet per second
Your key word is PER
Keep in mind that although there are set steps for working related rate problems but each problem is unique in its own method of solution.
To further explain the steps are guidelines but the specific application is dependent upon each problem.
The steps to solving these problems are as follows.
First, identify all given quantities and quantities to be determined. Also make a sketch and be sure to label the quantities.
Second, you must write an equation involving the variables whose rates of change either are given or are to be determined
Third, you must use the chain rule implicitly differentiate both sides of the equation with respect to time (T)
Finally, you must substitute all known variables into the resulting equation and solve for the desired rate of change
Be sure to print and study your Geometric formulas
Example:
A Pebble is dropped into a calm pond causing ripples in the form of concentric circles. The radius of the outer circle is increasing at a rate of 1 foot PER second when the radius is 4 ft at what rate is the total area (A) of the disturbed water changing?
#1 dr/dt= 1ft/sec
R=4 ft
dA/dt= ?
#2 A= pi r ^2
#3 dA/dt= pi{2r dr/dt}
#4 dA/dt= 2pi (4)(1)
Therefore 8 pi ft/sec

Blog

This week we learned implicit derivatives. Implicit derivatives are just regular derivatives with more than one variable so you have another variable to look at in the derivative. After you derive with your regular derivative rules, the only difference is the step afterwards. When you derive the second variable you have to show it is in respect to that variable, so you take your derivative and multiply by dy/dx or dt/dx then you solver for your dy/dx and that is your implicit derivative.

This was a homework problem from the week*

x^2+y^2=10

2x+2y(dy/dx)=0

2y(dy/dx)=-2x

dy/dx=-2x/2y

dy/dx=-x/y

Now you have your implicit derivative. This thing needs a word count. anyways, thats all i really know.

Week 6 Prompt

Is there an order of operations for the chain rule? Why or Why not? Give several examples of chain rule problems.

Blog for 10/3....not a pretty one

Honestly, I have no memory of last week in math
I can't remember anything we did, so I'm not gonna try to fool anyone
Mindsets go astray at times and you just forget everything
Sorryz >_>

well this week we learned about implicit derivatives. an implicit derivative is when you take a derivative with respect to another variable. for example, dy/dx would stand for the derivative of y with respect to the variable, x.

for example: 2y^2 - x^2 + 8x = 7

dy/dx[f(x)] = 4y - 2x + 8

you would then set dy/dx as a coefficient to y in order to solve for the implicit derivative.

(4y)(dy/dx) - 2x +8 = 0

all you do now is solve for dy/dx.

(4y)(dy/dx) - 2x = -8

4y(dy/dx) = 2x - 8

dy/dx = (2x - 8) / 4y

dy/dx = (x - 4) / 2y

Dustin Blog

This week we learned implicit derivatives. Implicit derivatives are very easy if you know your derivative rules. Implicit derivatives are just regular derivatives with more than one variable. What this means is you derive it with respect to another variable. After you derive with your regular derivative rules, the only difference is the step afterwards. When you derive the second variable you have to show it is in respect to that variable, so you take your derivative and multiply by dy/dx or dt/dx etc. After deriving, you solver for your dy/dx and that is your implicit derivative. Now this may seem complicating in words, but lets see a few examples.

x^2+y^2=10

First step is derive the equation using the normal derivative rules.

2x+2y(dy/dx)=0

I added the dy/dx to show that the derivative is in respect to y. The 10 became a 0 because you must derive both sides of the equation just like anything in algebra and due to the constant rule the derivative of 10 is zero. The next step is to solve for dy/dx using basic algebra.

2y(dy/dx)=-2x

dy/dx=-2x/2y

dy/dx=-x/y

Therefore -x/y is your derivative. If a point was given you could plug in at this time. If it asked for a second derivative of this equation though, the steps would be quite different. You take your first derivative and start to derive it like normal, but you plug in your first derivative where it occurs. And i guess thats about it about implicit derivatives. They do have word problems, that require some thinking outside of the box, but i don't have my book for examples and I really don't feel like making up a word problem at the moment. Good night.

Blog #6--Stephen

Hope everybody had a good weekend, I know I did. This week was pretty easy, I'm glad it's over, but I am ready for Spirit Week 2010! Anyway, this week in Calculus we learned about implicit derivatives.

When you "take" an implicit derivative, you are "taking" the derivative with two variables. However, the steps do not change when finding the implicit derivative. Work it just like a regular derivative.

d/dx is a derivative with respect to x. You use this when you take the derivative of x. This rule applies to d/dy also. The same concept applies.

When you take a derivative of anything but an x you write dy/dx, dr/dx, ds/dx, the top letter is your variable.

Gather all dy/dx and solve for dy/dx.

Example:

y^3 + y^2 - 5y - x^2 = -4

3y^2(dy/dx) + 2y(dy/dx) - 5(dy/dx) - 2x = 0

3y^2(dy/dx) + 2y(dy/dx) - 5(dy/dx) = 2x

dy/dx(3y^2 + 2y - 5) = 2x

dy/dx = (2x)/(3y^2 + 2y - 5)

dy/dx = (2x)/((3y+5)(y-1))

FINAL ANSWER

It's gonna take a while for me to get used to this complex derivative stuff, but hey, that's what Calc is, COMPLICATED. Oh well, peace out.

Blog #6 - Mary Graci

This week we learned in Calculus how to take the derivative of an equation with two different variables, called implicit derivatives. First of all, when referring to the derivative of this kind of equation, you use d(insert other variable)/dx. If the equation is x + y = 2, then the derivative would be called dy/dx. If the equation is x – r = 6, then the derivative would be called dr/dx.

Ex. 1) x + y = 2
First, take the derivative of each individual term, including what’s after the equal sign. When taking the derivative of the second variable (y), write “dy/dx” after the derivative.
1 + dy/dx = 0
Now, solve for dy/dx.
dy/dx = -1
And that’s your derivative!

Ex. 2a) Take the implicit derivative of: x^2 – 3y^2 = 6
2x – 6y dy/dx = 0
-6y dy/dx = -2x
dy/dx = 2x/6y
dy/dx = x/3y

We also learned how to take a second implicit derivative. It’s basically the same concept as taking the second derivative of a normal equation. You take the derivative (as shown above), then take the derivative again, then plug in the first derivative for dy/dx, and solve. If when you are finished you can plug in your original equation into your second derivative to simplify the answer more, do that and that’s your answer.
*a second implicit derivative is represented by d^2y/dx^2

Ex. 2b) Now take the second implicit derivative of example 2.
Use the quotient rule.
d^2y/dx^2 = ((3y)(1) – (x)(3dy/dx)) / 9y^2
= (3y – 3x dy/dx) / 9y^2
= (y – x dy/dx) / 3y^2
= (y – x (x/3y)) / 3y^2
= ((3y^2/3y) – (x^2/3y)) / 3y^2
= ((3y^2 – x^2) / 3y) / 3y^2
Sandwich it.
d^2y/dx^2 = (3y^2 – x^2) / 9y^3
Be sure to simplify the answer as much as possible.

Blog 6

Were stilllll on chapter two, and this week we learned about implicit derivatives. An implicit derivative is derivatives with two variables. The steps for a derivative and implicit derivative do not change. It simply means “with respect to”. For example d/dx- the derivative would be with respect to x. An example to solve a simple one: d/dx (x^3)= 3x^2. When you take a derivative of anything but an x, you write dy/dx, ds/dx, dr/dx **the top letter being your variable. You would gather all dy/dx and solve for dy/dx. So basically you take the derivative like usual, instead when you take the derivative of y you would have to put dy/dx behind it. After that you would combine like terms (if any), then solve for dy/dx. **Product, Quotient, and Chain rule may be used in implicit derivatives.

Example:
X^2+y^2=9

*First thing you would do is take the derivative. (Remember the derivative of all constants=0)
2x+2y(dy/dx)=0
*Next solve for y, so bring 2x to the other side
2y(dy/dx)=-2x
*Then divide by 2y
Leaving you with: dy/dx=x/y

**Now if you would have a point, you would just plug it in to the final answer. Say the point was (3,5)
The answer would then become: 3/5

**Nowww, if you would have to find the second derivative you would have to take the final answer which was x/y and solve for it. In this case you would use quotient rule. Keep in mind that when solving in again you still have to put dy/dx when taking the derivative of y.

Y(-1)-(x)(1dy/dx)/y^2
-y-x(dy/dx)/y^2
**From here you can plug in the original dy/dx
-y-x(x/y)/y^2 = -y-x^2/y/y^2
*Sandwich it and the answer would be –y^2-x^2/y^3
------------------------------------------------------
Now for something I don’t understand. I don’t understand the last thing we learned with all the word problems. When we do them in class I understand them. But if I would try one on my own I would be lost.

Alaina's blog, 3 Oct. 2010

This week in Calculus, we learned implicit derivatives. Basically it is just taking the derivative, but instead of having only one variable, “X”, there are two variables, “X” and “Y”.
Implicit Derivatives are represented as dy/dx with respect to X, or the bottom variable.
To take an Implicit Derivative:
You treat it just like a regular derivative, taking the derivative of each element. When you take the derivative of “Y”, you put dy/dx behind it. Then, you solve for dy/dx. Normally, you end up with a fraction.

Ex 1: X(squared) + Y(squared)=25
2X+2Ydy/dx=0
2Ydy/dx=-2X
dy/dx=-2X/2Y

Ex 2: 2X(squared)+Y(squared)=36X
4X+2Ydy/dx=36
2Ydy/dx=36-4X
dy/dx=(36-4X)/2Y
dy/dx= (18-2X) /Y

Also, we learned Implicit Second Derivatives. Implicit Second Derivatives are represented as
d(squared)y/dx(squared). Basically, you start with a function and follow the rules for regular Implicit Derivatives, which is your first derivative. From there, you take the derivative of your first derivative, second derivative, and plug the first derivative in for dy/dx and simplify.

Ex 1: X(squared) +2Y(squared) =4
2X + 4Ydy/dx =0
4Ydy/dx = -2X
dy/dx = -2X /4Y
dy/dx = -X /2Y
Implicit Second Derivative=> ((2Y)(-1)-[(-X)(2dy/dx)])/(2Y)(squared)
( 2Y+2Xdy/dx)/(2Y)(squared)
(-2Y+2X(-X/2Y))/(2Y)(squared)
(-2Y-(2X(squared)/2Y))/(2Y)(squared)
((-4Y(squared)- 2X(squared) )/2Y)/(2Y)(squared)
(-4Y(squared)-2X(squared )/(4Y)(cubed)
(2( -2Y(squared)-X(squared)))/(2(2Y)(cubed))
(-2Y(squared)-X(squared))/((2Y)(cubed))
X(squared)/Y(cubed)

Taylor blog #6

This week we learned about implicit derivatives with two variables.
The key thing to remember when taking implicit derivatives is that the steps are the same as taking a regular derivative.
“with respect to” is a phrase you will see often when dealing with implicit derivatives.
The phrase “with respect to” gives you a clue as to what you will be solving for
When you see “d/dx” you also must be able to recognize that this is a symbol for “with respect to x”
When you take a derivative of anything besides x you will write it in the asterisk spot of the following example: D*/Dx
In other words, the top letter next to the d is the variable you are taking the derivative of
For example:
With respect to y: dy/dx
With respect to r: dr/dx
With respect to s: ds/dx
To solve a problem you must follow three general steps:
First take the derivative
Second gather all d*/dx
Third solve for d*/dx
Example:::
Sqrtxy= x^2y+1
(xy)1/2=x^2y+1
X^ 1/2 y^ ½ = x^2 y +1
First take the derivative
X^1/2 (1/2y^-1/2 dy/dx) + y^1/2 (1/2x^-1/2)= x^2 (1dy/dx+y(2x))
1/2x^1/2y^-1/2 dy/dx + 1/2x ^-1/2 y^1/2 = x^2 dy/dx + 2xy
Second gather all d*/dx
1/2x^1/2y^-1/2 dy/dx- X^2 dy/dx= 2xy - 1/2x^-1/2y^1y^1/2
Third solve for d*/dx

Dy/dx = 2xy -1/2x^-1/2y^1/2 / ½ x^1/2 y^-1/2 – x^2

Example with trig:
4cosx siny=1
4[cosx(cosy dy/dx)+ siny (-sinx)] = 0
4cosxx cosy dy/dx – 4 siny sin x =0
4 cosx cosy dy/dx= 4siny sinx
Dy/dx= 4 siny sinx / 4 cosx cos y
Dy/dx= tan y tan x
We also learned how to take the second derivative when dealing with implicit derivatives
The steps to solving the second derivative of an implicit derivative are:
Take the first derivative
Solve for dy/dx
Take the derivative of dy/dx (this will be d^2y/dx^2)
Plug in the result of when you solved for dy/dx into the d^2y/dx^2 equation for dy/dx
Solve again
**** sometimes when you solve the numerator will turn out to be your original equation, if so, plug in what the original equation equaled and simplify.
If not then d^2y/dx^2 will be what you found when you “solved again”
Example:
x^2 + y^2 = 25
Take the first derivative
2x + 2y dy/dx = 0
Solve for dy/dx
Dy/dx= -2x/2y therefore -x/y
Take the derivative of dy/dx
- Y(1)- [ x(1dy/dx)]/y^2
- Y-x dy/dx / y^2
Plug in the result of when you solved for dy/dx into the d^2y/dx^2 equation for dy/dx

- Y-x ( -x/y)/ y^2

Solve again

- Y^2 + x^2/y/y^2/1
-y^2 +x^2/ y^3
**** remember that sometimes when you solve the numerator will turn out to be your original equation, if so, plug in what the original equation equaled and simplify.

Therefore:
-25/y^3

Devin's Reflection

Guidelines for Implicit Differentiation

1. Differentiate both sides of the equation with respect to x

2. Collect all terms involving dy/dx on the left side of the equation and move all other terms to the right side of the equation

3. Factor dy/dx out of the left side of the equation

4. Solve for dy/dx

To find an Implicit Differentiation

1. Differentiate both sides of the equation with respect to x

2. Collect the dy/dx terms on the left side of the equation and move all other terms to the right side of the equation

3. Factor dy/dx out of the left side of the equation

4. Solve for dy/dx by the left side of the equation.

Determine the slope of the tangent line to the graph of 2x^2+y^2=2

Ex. First take the derivative of the equation

4x+2ydy/dx=0

Solve for dy/dx

Dy/dx= -4x/2y

Dy/dx= -2x/y

Trig and Differentiable Function

Find dy/dx implicity for the equation sin y=x. Then find the largest interval od the form –a <>

Ex. d/dx[tany]=d/dx[x]

Sec^2ydy/dx=1

Dy/dx=1/sec^2y

10/2/10

This week we learned implicit derivatives, derivatives with two variables (x and y) and derivatives "with respect to" a variable, such as dy/dx >> means to take the derivative with respect to y. That goes for any variable (where the top letter is your variable). So keeping that in mind, if you have an equation like x^2+y^2=0, you take the derivative all the way across, but for the y^2 you get 2y dy/dx..because you're taking the derivative with respect to y. This goes for anytime you see a y in an eqation. And remember when you take an implicit derivative, you are solving for dy/dx!!!!!!!!!!..It's pretty easy, except there's a lot of algebra involved, so I think I could used more practice with that. But anyway, here are some examples of what we did this week:

Ex. 1) x^3y^3 - y = x
*The first thing you should notice is that this will be an implicit derivative (duh) because there is both x's and y's in the equation
*The next thing you should notice is that there's a product rule also. x^3y^3..so you should take care of that first
*So you get:
(x^3)(3y^2dy/dx)+(y^3)(3x^2)-dy/dx = 1
*Now you can simplify what you just got and you get
3x^3y^2dy/dx + 3x^2y^3-dy/dx = 1
*Now you have to move the dy/dx's to one side of the equal sign and everything else to the other side
*So you end up with
3x^3y^2dy/dx-dydx = 1-3x^2y^3
*Now you have to solve for dy/dx, so first you would factor out a dy/dx then, you would divide by 3x^3y^2-1
*So your final answer is:
dy/dx = 1-3x^2y^3/3x^3y^2-1

Ex. 2) x^(2/3) + y^(2/3) = 5 (8,1)
Find dy/dx by implicit differentiation and evaluate the derivative at the given point
*First, you start off by taking the derivative like any other problem..putting dy/dx after you take the derivative of the y^2/3..So you have:
2/3(x)^(-1/3) + 2/3(y)^(-1/3)dy/dx = 0
*Now you can simplify and move the 2/3x^(-1/3) to the other side at the same time
*So you have 2/3y^(1/3)dydx = -2/3x^(1/3)
*Finally you divide by 2/3y^(1/3) and you would have to sandwich it..
So after you sandwich it you get -6y^(1/3)/6x^(1/3) and you can cancel out the 6's to make your life easier.
*So you have -y^(1/3)/x^(1/3)
*Unfortunately, you're not done yet. You still have to plug in the point they gave you.
*So you get
-(1)^(1/3)/(8)^(1/3)
= -1/2

Ex. 3) Find an equation of the tangent line of the graph at the given point
(y-3)^2 = 4(x-5) (6,1)
*Firstttttttttt, take the derivative all the way across. **Notice there's a chain rule!
So you get:
2(y-3)(dy/dx) = 4(1) **Remember, leave constants, like 4, out in front
*Now to solve for dy/dx, all you have to do is divide by 2(y-3)
*So you have (4)/2(y-3)...and you can factor out a 2 from the numerator and denominator
*So you get that dy/dx = 2/y-3
*Now you just plug in the point they gave you and you get:
2/-2
=-1 and that's your slope for the equation you're about to set up
*using the point slope formula, like always, you end up with:
y-1=-1(x-6)

**As for what I didn't understand...I'm going to go out on a limb here and say I am LOST with those word problems! (well not totally, like I understand them when Ms Robinson explains them, but I'm not sure I'll be able to work through one by myself just yet) And I could use some more practice with taking the 2nd derivative of implicit derivatives..my algebra is killing me..Other than that, I'd say I'm doing pretty good