Week 5 Prompt

How do you find the equation of a tangent line? how does it differ from a regular derivative. Give an example of finding an equation of a tangent line and an example of finding a regular derivative. Include the directions that would be given with each problem.

Next Blog (can't remember #)

Seriously, I'm gonna fail the test tomorrow
I can't remember anything about these damn derivatives
I'm seriously at a blank state of mind about math right now and I don't know what to say or put on this blog
At least I'm being honest and not copying anybody or anything
:/

Devin's Blog

Derivative is a slope of a tangent line, slope of a curve at a point, or a rate of change. If an equations states something about a rate, a velocity, a speed, instantaneous velocity, or an acceleration then you must find a derivative. When using the constant rule, you must take the derivative of the equation and then set the derivative equal to zero. When using the power rule, every time you take a derivative, you must lose a power. For composite functions you must use the chain rule. When using the chain rule you work from outside to the inside. You must take the derivative of the outside and then recopy the inside of the equation and continue with the problem. You can also use the chain rule inside of using a product or quotient rule.

Ex. y’ 6x^4

4(6)x^4-1

=24x^3

Ex. 5x^2+11x

2(5)x^2-1+1(11)x^1-1

=10x+11

Ex. x^5+3x

5(1)x^5-1+1(3)x^1-1

=5x^4+3

I could use a lil help on how to do a product and chain rule though.

Justin 5

This week we learned the chain rule, which if you miss can get pretty confusing. You use it when more than one function is going on (yo dawg we heard you like functions).
ex: sin(cos(x^2
The formula:

f'(g(x))x g'(x)

Derivative of the outside
Recopy the inside
Multiply by the derivative of the inside

(x^2+3)^3
Outer:y=u^3
Inner:u=X^2+3
3(x^2+3)(2x)
3(2x)(x^2+3)

Im having trouble with trig functions and "scary" looking problems

This week we learned the chain rule and i missed it and that kind of KILLED ME. I know how to do it now kind of so ill explain it. The formula for the chain rule is f’(g(x)) x g’(x). –Derivative outside, recopy inside, times derivative of inside. d/dx= [f(g(x)]

examples:
y=(5x-8)^4 The outer would be y=u^4, and the inner would be u=5x-8
Y=csc^3x The outer would be y=u^3, the inner would be u=csc x
Y=tan^2x

-First thing you do is bring the exponent to the front
2 tan x
-Then you have to multiply it by the derivative of tan
2 tan x x (sec^2x)
The answer would be 2 tanxsec^2x

I need to learn how to do the parallel and perpendicular slope problems.

blog 5

well this week we learned chain rule. im not absolutely sure im doin it right but im givin it a shot.

the formula for chain rule is: dy/dx [f(g(x))] = f ' (g(x)) * g ' (x)

for example: 3x + 2tan(3x)^2

3 + (4sec^2(3x))(18x)

3 + 72sec^2(3x)

and i think thats all...

as a side note, we also learned quotient rule

the formula for quotient rule is: dy/dx [f(x)/g(x)] = ((g(x))(f '(x)) - (f(x))(g'(x))) / (g (x))^2

Blog 5

Were still on chapter two and each section gets a little harder. I thought it was pretty easy until we got to the product rule mixes into the chain rule. Anyways, I will explain chain rule because I understand it okay. The formula for the chain rule is f’(g(x)) x g’(x). –Derivative outside, recopy inside, times derivative of inside. d/dx= [f(g(x)] **The chain rule can also be called the general power rule.
In order to be able to do these problems you need to understand what is the outer and inner of the problem (I have a hard time with this myself.) The first few exercises of our homework problems were like this.

A few examples of identifying outer and inner:
y=(5x-8)^4 The outer would be y=u^4, and the inner would be u=5x-8
Y=csc^3x The outer would be y=u^3, the inner would be u=csc x

Now and example using the chain rule:
Y=tan^2x
-First thing you do is bring the exponent to the front (General Power Rule)
2 tan x
-Then you have to multiply it by the derivative of tan
2 tan x x (sec^2x)
The answer would be 2 tanxsec^2x

Something that I didn’t really understand was the word problems. I understand which ones are velocity, and average velocity, and I understand what you have to do to them. I just don’t get the complicated and more detailed ones. Like when something drops on impact, or when you have to plug in zero to the problem.

Blog #5

Ok so this week was pretty hard. I'm having a lot of trouble with 2.3 and 2.4 just how to use the quotient rule, and I never seem to get the answers that the book has. And I'm still not sure how and when to use the Chain Rule and what not. Anyways, I'll try to explain to the best of my ability, and hopefully study a lot tonight and do at least halfway decently on the test tomorrow. :/ Ok, to the math.

When you're working with the chain rule and composite functions, you work from the outside in.

The Chain Rule is defined as

d/dx[f(g(x))] = f ' (g(x)) * g'(x)

(derivative of the outside, recopy the inside multiplied by the derivative of the inside)

Example:

2 * 8th root of (9-x^2)

2 * (9-x^2)^(1/8)

2 * (1/8)(9-x^2)^(-7/8) * (-2x)

(-x/2)(9-x^2)^(-7/8)

(((-x)/(2)))/(((9-x^2)^(7/8))/(1))

((-x))/(2(9-x^2)^(7/8)) FINAL ANSWER

I already said what I didn't understand in the top of the blog. Some examples of Chain Rule and quotient rule used in different occasions would be nice, especially with trig functions.

9/26 Blog

This week we learned the chain rule and stuff. Its basically the same as the power rule just used at different times with one difference. We also learned the general power rule, which is a form of chain rule.

The chain rule is used when you have two things going on. For example:

(x^2+x+3)^2

First step is like normal, you would move the exponent to the front.

2(x^2+x+3)^2

Next, you subtract one from the original exponent to form the new exponent.

2(x^2+x+3)

The next step is to take the derivative of the inside and multiply by it.

2(x^2+x+3)(2x+1)

Then you factor/simplify if you can, but this one is as simplified as it needs to be.

So, that's about it for what we learned this week.

9/25/10

Alright, so we started this past week off learning the chain rule..which is a pretty easy concept, the only trouble is knowing exactly WHEN to use it. That's still getting me with some problems. But anyway, you would use the chain rule when there is something on the "inside" of a trig function..I guess you could say? And you would normally use it at times when you see parenthesis in an equation with an exponent on the outside..that would specifically be called the General Power Rule. Then later in the week we got back to some advanced math skills, solving trig equations again. That's not too bad either, I just basically needed to refresh my memory on it. Sooooo, here are a few examples of things we reviewed this week:

Ex 1) y = 2(6-x^2)^5 ..Find y'
*Okay so the first thing you should notice is the parenthesis and the exponent on the outside. You want to deal with the outermost thing FIRST, so you would start with the 2.
*Since the 2 is just a number you leave it out front..then the chain rule tells you to bring the exponent, 5, to the front and subtract 1 and recopy the middle 6-x^2, so you have: 2(5(6-x^2)^4
*The next part of the chain rule says to take the derivative of the middle, which is 6-x^2, and then multiply it by what you have so far. So the derivative of that would be -2x (since d/dx of 6, a constant, is zero)
*So now you have: 2(5(6-x^2)^4 X (-2x)
*Now you simplify. And you get -20x(6-x^2)^4 (Since there are no negative exponents, you will not have a fraction as your answer)

Ex 2) f(x) = tan^2(5x) ..Find the derivative.
*The first thing you should notice about this problem is that it is NOT a product rule, because you have more than just "tan x". You have something within the trig function, so you have to start with the outermost thing and then take the derivative
*To start off, I like to rewrite these kinds of problems, because it helps me see and recognize things easier. So I would rewrite it as [tan(5x)]^2 ..Because that's basically the same thing as the original function. So now you can easily see that the outermost thing is the exponent, so that tells you that you have to pull it out front and subtract 1, and recopy the middle, tan(5x).
*So that means you have: 2(tan(5x))
*Now you have to take the derivative of the inside, but only the first part, "tan". So once you take the derivative, you multiply it by what you already have...So the derivative of tangent is sec^2x. Then you recopy what's left in the middle, which is 5x
*So now you should have:
2(tan(5x)) X (sec^2(5x))
*Now you can take the derivative of 5x which is 5. And you multiply 5 by what is above ^^
*So you have 2(tan(5x)) X (sec^2(5x)) X (5)
*Finallyyyyyy, you simplify all of that!
*And you should get 10tan(5x)sec^2(5x)
***P.S.-- "X" means "times" here.

**As for what I didn't understand this week. Eh, it wasn't too bad, I'm JUST starting to grasp everything though, haha. (like using product rule, chain rule, quotient rule, etc. simultaneously) But hopefully more practice within the next 24 hours or so will help.

Taylor Blog #5

This week we learned the chain rule and the general power rule
Im going to explain the Chain Rule,
The chain rule could loosely be defined as an order of operations that is used when solving composite functions.
A very important key concept to remember when using the chain rule is to work from the outside to the inside.
The “Formula” for the chain rule is
d/dx [fg(x)] = f ’(g(x)) (g’(x))
The most common procedure to using the chain rule is
• First, take the derivative of the outside
• Second, recopy just the inside
• Finally, multiply by the derivative of the inside
Example:
Square root of 3x^2-X+1
= (3x^2-X+1)^1/2
First, take the derivative of the outside:
½(________)^-1/2
Second, recopy just the inside
½(3x^2-X+1)^-1/2
Finally, multiply by the derivative of the inside
½(3x^2-X+1)^-1/2 (6x-1)
Simplify
6x-1/2(3x^2-x+1)^1/2


Another problem that may come up on the AP Exam that I feel I should include
The problems where we fill out the given chart for
Y= F(g(x))
U= g(x)
Y= f(u’)

Example:
Y= F(g(x)) U= g(x) Y= f(u’)
Y=(5x-8)^4 U=5x-8 y=u^4

I just thought I’d include this example because it helps you to understand the composition of an equation.

Blog #5 - Mary Graci

Okay, this week we learned the chain rule. You use the chain rule when you have a function inside of a function, f(g(x)).

Ex. (x^2 + 3)^4
First, take the exponent of the entire function, bring it to the front, then subtract one from the exponent, and then just copy the inside.
4(x^2 + 3)^3
Now, multiply that by the derivative of the inside equation.
4(x^2 + 3)^3 (2x)
Put in simplest terms.
= 8x(x^2 + 3)^3

You can also use the chain rule with the product rule or quotient rule. If this is the case, then you first need to determine whether the chain rule is on the outside or the inside; this is to figure out what order you do things in. If the chain rule is on the outside, you do that first. If not, you do the product rule or quotient rule first.

Ex. (x + 3)(x^2 – 1)^2
In this problem, the product rule comes first. When taking the derivative of the second equation, use the chain rule.
(x + 3)(2(x^2 – 1)(2x)) + (x^2 – 1)^2 (1)
= 4x(x + 3)(x^2 – 1) + (x^2 – 1)^2

Ex. ((x + 2)/(x – 1))^2
For this problem, the chain rule is on the outside, therefore it goes first.
2((x + 2)/(x – 1)) (((x – 1)(1) – (x + 2)(1)) / (x – 1)^2))
2((x + 2)/(x – 1)) ((x – 1 – x – 2) / (x – 1)^2)
2((x + 2)/(x – 1)) ((-3)/(x – 1)^2)
= -3(2x + 4) / (x – 1)^3


Besides this, we just reviewed everything in Ch. 2 for the test Monday. ^^

Week 4 Prompt

What are the different keywords that indicate taking a derivative? Give examples of direction sets that tell you they want a derivative.

Blog 4

Ahh, this derivative stuff is still soo confusing, and these new rules aren't helping at all......
Can I shoot my Calculus book? Please? Pretty Please? With Sugar On Top? :D

I'm just gonna predict that you said no to that question...or rather...series of questions.....

Seriously, all I can remember/understand from last week is this:

dx/dy sinx = cosx
dx/dy cosx = -sinx

That's all I've got.

Alaina's blog, 19 Sept 2010

This week in Calculas AB we learned the difference between average velocity, velocity and instantaneous velocity. Average velocity implies finding the slope; instantaneous velocity means to take the derivative. Also, we learned product and quotient rules:

Product Rule:

[f(x)g(x)]=f(x)g¹(x)+g(x)f¹(x)

Quotient Rule:

[f(x)/g(x)]= g(x)f¹(x)-[f(x)g¹(x)]/g(x)²

The product and quotient rules are pretty simple. All you do is plug into them.

Ex 1. [(x²+4)(x-2)]
=(x²+4)(1)+(x-2)(2x)
=(x²+4)+(2x²-4x)
=3x²-4x+4

Ex 2. [(x²+4)/(x-2)]
=(x-2)(2x)-[(x²+4)(1)]/(x-2)²
=2x²-4x-x²-4/(x-2)²
=x²-4x-4/(x-2)²
=(x-4)(x+1)/(x-2)²



The last thing we went over was taking the 2^nd and 3^rd derivative; all you do is take a derivative two or three times. Second and third derivative are represented like say g-prime or g¹, but instead of 1, it is 11 or 111-> g¹¹ or g¹¹¹.

Ex 1. y¹¹[x²]
y¹=2x
y¹¹=2

Ex 2. y¹¹¹[x²]
y¹=2x
y¹¹=2
y¹¹¹=0

Blog #idk i missed a few

this week in calc i learned how to find a derivative using a shortcut. you can use this shortcut to lessen the length of time it takes to find the derivative of polynomials or if the number is a constant then the derivative is automatically zero.

you do this by taking the exponent of the variable and setting it in front of the variable then subtracting one from the exponent and setting it as the new exponent of the variable.

for example: f'(x)=x^5

f'(x)=nx^n-1

f'(x)=5x^4

blog 4

this week we learned about product and quotient rule of derivatives

Product Rule:
f(x)g'(x)+g(x)f'(x)

Quotient Rule:
g(x)f'(x)-f(x)g'(x)/(g(x))^2

the idea is its not all about your calculus work, its more about algebra after plugging in which is like a flashback on its own haha and it's also important that you simplify as much as possible to get the exact answer.

Product rule ex:
(3x-2x^2)(5+4x)
(3x-2x^2)(4)+(5+4x)(3-4x)
12x-8x^2+15+12x-20x-16x^2
Dx=-24x^2+4x+15

Quotient rule ex:
5x-2/x^2+1=(x^2+1)(5)-[(5x-2)(2x)]/(x^2+1)^2
5x^2+5-[10x^2-4x]/(x^2+1)^2
5x^2+4x+5/(x^2+1)^2

9/19/10

Soooo, time for another blog. Starting this week off we learned the difference between average velocity, velocity, and instantaneous velocity. For instance, when you see the words average velocity that means to find the slope. When you see the words instantaneous velocity (or if it's implied) then that means to take the derivative. Then we learned the product and quotient rules used to take derivatives for insanely large functions. I thought that was the easiest thing we learned this week, because all you have to do is remember the formula. On the other hand, some problems get really tricky when it comes down to the simplifying. And also, we learned the derivatives of more trig functions. So here are a few examples of some things we did this week:

Ex 1) f(x) = -1/x [1,2]
^Find the average rate of change of this function over the given interval.
*Alright well first you should notice the key words, average rate of change. This is telling you that you need to used the slope formula
*So the first thing you should do is take your two x-values, 1 & 2, and plug them separately into the f(x) function to get their y-values
*So when you plug in 1 you get: -1/1 = -1
*And when you plug in 2 you get: -1/2 = -1/2
*Then you can plug those numbers into the slope formula y2-y1/x2-x1
*So you get (-1/2 + 1)/2-1 = (1/2)/1
*= 1/2

Ex 2) f(x) = x^3cosx
^Differentiate using the product rule
*Okay so the first thing you should do is decide what two things are being multiplied.....and that would be x^3 and cosx. So you can say x^3 is "f" and cosx is "g"
*Using this formula >> f(x)g'(x) + g(x)f'(x) ..take the derivative
*So this is what you would get when you plug in:
(x^3)(-sinx) + (cosx)(3x^2)
*Then you would distribute that out and get this:
-x^3sinx + 3x^2cosx
*Now you ask yourself if there's anything else you can do to further simplify your answer..
YES!..You can take out a -x^2 to make your life a little bit easier
*And you get:
d/dx = -x^2(xsinx - 3cosx)

Ex 3) f(x) = (x^2-4)/(x-3) c=1
^Differentiate using the quotient rule.
*Sooo first you can label your numerator as "f" and the denominator as "g" to make it easier for you to plug into the formula
*Plugging into this formula >> g(x)f'(x) - f(x)g'(x)/[g(x)]^2 ..you get:
(x-3)(2x) - [(x^2-4)(1)]/(x-3)^2
*Next you can distribute the numbers through. **Don't forget to distribute the negative!
And you get:
2x^2-6x-x^2+4/(x-3)^2
*Now you can combine the like terms in the numerator and you end up with this:
x^2-6x+4/(x-3)^2
*Then you see if there's anything you can do to simplify it further....anddddd you can't so now you can plug in "c" (1) to get your final answer
*So you get: 1^2-6+4/(1-3)^2
*= -1/4

**Okay here's what I didn't quite grasp this week...I could use some further explanation on those velocity word problems, I honestly wouldn't know what to do with them except stare them down on a test paper..and that would probablyyyyyy waste alot of time so I think it would be better if I completely understood how to work them. :)

Blog #4

Well, this week went by pretty fast, and we are all still on Chapter 2. But section 2.3. I really don't understand the quotient rule that we just learned this week...I mean I understand how to use it but I'm having some problems with the actual plugging in and solving part of it. Anyways, I'll try to explain to the best of my ability the product and quotient rules. You have to use f(x) and g(x) in these formulas btw. The asterisk is a multiplication symbol btw. This is confusing btw.

Product Rule

d/dx [f(x)*g(x)] = f(x)*g'(x) + g(x)*f '(x)

Quotient Rule

d/dx [(f(x))/(g(x))] = (g(x)*f '(x) - f(x)*g'(x))/([g(x)]^2)

I'll give an example of the product rule, since I'm kinda rusty on the quotient rule.

Find the derivative of the product of (2x-x^2)(4+2x)

(2x-x^2)(2) + (4+2x)(2-2x)

Just FOIL this out.

4x - 2x^2 + 8 - 8x + 4x - 4x^2

Comes out to a simplified form:

-6x^2 + 8

That is the derivative, using the product rule.

As I said above, I am really bad on the quotient rule, so a few examples would be really nice lol. Thanks.

Blog 4?

Were still on chapter two and so far it is pretty easy. Well the concept of it is pretty easy; you just have to be careful when doing the algebra. I will explain product and quotient rule from 2.3 because it is very simple. For both of the rules you just have to plug in to the formula, so you would have to have the formulas memorized because for each rule there is a different formula. Keep in mind the derivative of sin x= cos x, and the derivative of cos x=-sin x.
Product Rule:
d/dx [f(x)g(x)]=f(x)g’(x)+g(x)f’(x)
Quotient Rule:
d/dx (f(x)/g(x))= g(x)f’(x)-f(x)g’(x)/[g(x)]^2
**f’(x) and g’(x) means to find the derivative, for the ones that are just f(x) or g(x) you plug in the normal number.
Ex: (3x-2x^2)(5+4x)
*First thing you are going to do is plug into the product formula.
(3x-2x^2)(4)+(5+4x)(3-4x)
*From here you are going to foil it out
=12x-8x^2+15-20x+12x-16x^2
*Combine like terms
Dx= -24x^2+4x+15 would be your final answer (Dx->Derivative)

Ex: 5x-2/x^2+1
*Plug into the quotient formula
(x^2+1)(5)-(5x-2)(2x)/(x^2+1)^2
*Foil it out
5x^2+5-10x^2+4x/(x^2+1)^2
*Combine like terms
-5x+4x+5/(x^2+1)^2
*From here you are going to try and factor it any way possible. Quadratic formula, factoring, pulling numbers out, completing the square, any way possible. You are going to realize that this problem can’t be factored so it would be left as it is.
dy/dx= -5x^2+4x+5/(x^2+1)^2 (would be the final answer)
Something I think is a little difficult is still in section 2.3. It’s with tan and cot and all that. I’m not sure how to do it really, but I don’t really know all the trig functions.

Devin's Blog

The informal of definition of a limit is “what is happening to x as x gets close to a certain number.” In order for a limit to exist, we must be approaching the same y-value as we approach some value c from either the left or the right side. If this does not happen, we say that the limit does not exist as gets closer to c.

If we want the limit of f’x as we approach some value of c from the left hand side, we will write

The concept of limits as x approaches infinity means the following: “what happens to x as x gets infinitely large.”

We are interested in what is happening to the y-value as the curve gets farther and farther to the right. We can also talk about limits as x approaches negative infinity. This means what is happening to the y-value as the curve gets farther and farther to the left. The terminology we use are the following:

Although we use the term “as x approaches infinity”, realize that x cannot approach infinity as infinity does not exist. The term x approaches infinity” is just a convenient way to talk about the curve infinitely far to the right.

There are only 4 possibilities for a limit the curve can go up forever. In that case, the point on the curve can go down forever. In that case, limit does not exist. For convenience sake, the limit does not exist.

blog 4

This week we learned about the product rule and the quotient rule of derivatives
For each of these things you need to use the exact formula needed for the equation. If you do the equation wrong then your answer will be wayyyyy off.
The product rule:
The product rule is recognized as F(x)G(x)
The formula for solving with the product rule is
D/Dx [F(x)G(x)] = F(x) Gprime(x) + G(x) Fprime(x)
First I take the derivative e of each so that when I plug in I already know what the derivatives are to be plugged in
The quotient rule looks like F(x)/G(x)

There might be several types of equations in each problem
The formula for solving with the quotient rule is

D/Dx [F(x)/G(x)] = G(x) Fprime(x)- F(x) Gprime(x)/ [G(x)]^2
When you are solving with the quotient rule the negative must be distributed to the entire product of f(x)G prime(X)

D/Dx [3x-2x^2]= 3-4x
Therefore F prime= 3-4x
D/Dx [5+4x]= 4
Therefore G prime= 4
Now you plug into the formula
3x-2x^2(4)+ 5+4x (3-4x)
12x^2x-8x^2+15-20x+12x-16
-24x^2 + 4x +15
and your answer is
Dx= -24x^2+4x+15

Example: 5x-2/x^2-1
First I take the derivative of each so that when I plug in I already know what the derivatives are to be plugged in
D/dx [ 5x-2] = 5
Therefore F prime= 5
D/dx [x^2 – 1] = 2x
Therefore G prime = 2x
Now you plug into the formula
(x^2+1)(5)-[(5x-2)(2x)]/(x^2+1)^2
5x^2+5-[10x^2-4x]/(x^2+1)^2
5x^2+5-10x^2+4x/(x^2+1)^2
-5x^2 + 4x+5/ (x^2+1)^2
So now your final answer is.
DX= -5x^2 + 4x+5/ (x^2+1)^2

Blog #4 - Mary Graci

This week, we learned the product rule and quotient rule for taking a derivative. Each rule follows a formula and all you have to do is plug in the equations into the formulas when necessary and work it out.

The product rule formula: f(x)g’(x) + g(x)f’(x)
When using the product rule, the problem must state that you take the derivative of two separate equations multiplied by each other.
**note: g’(x) or f’(x) means take the derivative of g(x) or f(x)

Ex. 1) (x+3)(x^2+6x)
(x+3)(2x+6) + (x^2+6x)(1)
2x^2 + 6x + 6x + 18 + x^2 + 6x
3x^2 + 18x + 18
= 3(x^2 + 6x +6)
**note: make sure the answer is as fully simplified as possible

The quotient rule formula: (g(x)f’(x) – f(x)g’(x)) / (g(x))^2
When using the quotient rule, the problem must state that you take the derivative of two equations divided by each other.
**note: if the denominator is just a single term, it is not considered an equation and you do not use the quotient rule; you simply rewrite the problem and take the derivative normally

Ex. 2) (x^3+4) / (x-2)
((x-2)(3x^2) – (x^3+4)(1)) / ((x-2)^2)
((3x^3-6x^2) – (x^3+4)) / ((x-2)^2)
**note: to keep things in simplest terms, do not factor out the bottom
(3x^3 – 6x^2 – x^3 – 4) / ((x-2)^2)
**note: be sure to distribute the negative throughout the second equation
(2x^3 – 6x^2 – 4) / ((x – 2)^2)
(2(x^3 – 3x^2 – 2)) / ((x – 2)^2)

We also learned that in word problems, if it says to find the velocity or the instantaneous velocity, it means to take the derivative of the position function. If it says to find the average velocity, it means to find the slope by plugging in the two x-values into the position function to get the two y-values and using the slope formula (y2 – y1) / (x2 – x1).


Anyway, I pretty much understood everything this week. Now, I wasn't at school Thursday, so I missed that lesson, but I'm sure I'll catch up.

Taylor Blog #4

This week we learned about the product rule and the quotient rule of derivatives
For both the product rule and the quotient there is a recognizable format that will allow you tho know which rule you will need to use and for each rule there is a formula to memorize and put into effect to find the derivative.
The product rule:
The product rule is recognized as F(x)G(x)
The formula for solving with the product rule is
D/Dx [F(x)G(x)] = F(x) Gprime(x) + G(x) Fprime(x)
Example: (3x-2x^2) (5+4x)
First I take the derivative e of each so that when I plug in I already know what the derivatives are to be plugged in
D/Dx [3x-2x^2]= 3-4x
Therefore F prime= 3-4x
D/Dx [5+4x]= 4
Therefore G prime= 4

Now you plug into the formula
Therefore
3x-2x^2(4)+ 5+4x (3-4x)
Distribute
12x^2x-8x^2+15-20x+12x-16
Simplify
-24x^2 + 4x +15
Because this equation cannot be simplified any further
Dx= -24x^2+4x+15


The quotient rule:
The quotient rule is recognized as F(x)/G(x)
The formula for solving with the quotient rule is
D/Dx [F(x)/G(x)] = G(x) Fprime(x)- F(x) Gprime(x)/ [G(x)]^2
Example: 5x-2/x^2-1
First I take the derivative of each so that when I plug in I already know what the derivatives are to be plugged in
D/dx [ 5x-2] = 5
Therefore F prime= 5
D/dx [x^2 – 1] = 2x
Therefore G prime = 2x
Now you plug into the formula
Therefore
(x^2+1)(5)-[(5x-2)(2x)]/(x^2+1)^2
Distribute
5x^2+5-[10x^2-4x]/(x^2+1)^2
Distribute the negative
5x^2+5-10x^2+4x/(x^2+1)^2
Simplify
-5x^2 + 4x+5/ (x^2+1)^2
Because this equation cannot be simplified any further
DX= -5x^2 + 4x+5/ (x^2+1)^2

Some things to remember:
**don’t forget that when solving with the quotient rule the negative must be distributed to the entire product of f(x)G prime(X)
** don’t forget which order F(X)G prime(X) and G(X)F prime(X) go in for each equation because it DOES matter.

Blog for 1/16

Well, this week we learned a little bit more about derivatives.....sadly, I still can't do them too well......so I think I'll just talk about stuff being differentiable.......

differentiability is the ability to take a derivative from a function

there are many ways that a function cannot be differentiable
those ways are:

1) if there is a curve
curves occur when absolute values and/or piecewises are found in the function that you are trying to take the derivative of
an example of a function with a curve is:
1) |3-x| you set what's inside of the absolute value equal to 0 and solve for x, so |3-x| is not differentiable at x=3

2) if there is a vertical asymptote
vertical asymptotes occur when x is raised to the (odd/odd)
x^5/9 would be a vertical asymptote, so x would not be differentiable at x=1

3) if it's not continuous
if a function is not continuous, then it's not differentiable, it's as simple as that

4) if there's a cusp
a cusp is when you have x^(even/odd)
x^2/7 would not be differentiable at x=0

Week 3 Prompt

If a graph is continuous is it differentiable? Why or Why not? Give examples to support your conclusion. If a graph is differentiable is it continuous? Why or Why not? Give examples to support your conclusion.

Devin's Blog

This week in class we have been learning about differentiability. This was a somewhat difficult to understand. We learned that taking a derivative is taking a slope of a function.

Differentiability

A function is not differentiable if:

1. There is a corner

a. Absolute value or a piecewise

2. Vertical tangent line

a. X^odd/odd

3. If it is not continuous

a. Jumps

b. Removables

c. Vertical asymptotes

4. There is a cusp

a. X^even/odd

Examples:

a. x+2/(X^2)

=there is a corner

b. X^5/3/(x)

=there is a vertical tangent line

c. X+1/x+1

=the function is not continuous

d. X^4/3/(x)

=there is a cusp

-If a function is not differentiable at a point you can take a derivative, but you cannot plug in “x”.

Using the steps and skills above you can educationally answer the following questions.

If a function is differentiable, is it continuous?

Yes

If a function is not differentiable, is it continuous?

Sometimes

If something is not differentiable, does it have a tangent line at that point?

No

Blog #3

SHORTCUTS! Yea, so this week we found out how to apply some derivative shortcuts. I'm one of those people who most likely will screw up the shortcuts more than the long formula, but it helps on timed tests. There are many different shortcuts. There is the constant rule, the powers rule, and many others.

The constant rule is by far the easiest of them all. It states that the derivative of any constant is 0.
Ex. Find the derivative of 10.
The answer would be zero because of the power rule.
To prove this i will plug it in to the definition of a derivative formula.

f(x-dx)-f(x)/dx

10-10/dx

0/dx

0

The other shortcut i mentioned is the power rule. This one is pretty cool. What it says is that when you have any variable with an exponent, to find the derivative you can follow these simple steps.

1) Move the exponent to the front of the equation.
2) Subtract 1 from the original exponent for the new exponent.

Ex. Find the derivative of x^4.

x^4

4x^3

The 4 moved to the front. Then I subtracted one from the original exponent(4) and got 3 as my new exponent.

Blog #3

I like the fact derivatives have a shortcut. when i found out how easy it was it blew my mind (sorta). anyway we worked with derivatives again this week, as well as some new problems that involve them. here are some new rules we learned with derivatives

the power rule:
when taking the derivative of something, you take off a power.

formula: X^n = nX^n-1

ex: x^3 + 2x = 3x^2 + 2

the constant rule:
if there is a constant, its derivative is 0

ex: f(x)=3 f`(x)=0

SIN and COS:
basically you switch them up, but the derivative for cos=-sin

ex: -cosx = sinx

Taylor Rodriguez Blog #3

Toward the end of this week we focused in the shortcut to finding a derivative.
There are two rules to follow when using the derivative shortcut
These rules are : * The Constant Rule
• The Power Rule
The constant rule states that the derivative of a constant is 0
Therefore when asked to find a derivative of an equation with a constant the constant will automatically become 0.

The Power rule states that every time you take a derivative you lose a power.
Therefore if you are not taking the derivative of a constant you bring the exponent to the front of the variable and subtract one from the exponent.
The formula that displays this is
d/dx [x^n ] = nx^n-1
Example: d/dx[x^3] = 3x^2

There are many variations of equations that will call for simplification before you can take the derivative

When taking the derivative of a fraction you must simplify the equation by taking the numerator and placing it in front of the x and the exponent behind the denominator will become a negative
Example: d/dx [1/x^2] = x^-2
Then you would proceed to use the shortcut method to find the derivative.

When taking the derivative of a root you will turn the value of the root into a fraction with the exponent of x over the value of the root.
Example: d/dx [cuberoot x] = X^1/3
Then you would proceed to use the shortcut method to find the derivative.
**do not forget that after finding the derivative of a simplified equation you must convert the derivative back to an unsimplified form
Example: d/dx [ x^1/3] = 1/3x^-2/3 = 1/3cuberoot x^2

Stephen's Blog #3

Okay, so I loved this week because we only had 4 days! Yay. I would like to say that I actually like working with derivatives. They are pretty easy to me, as far as we got on them so far at least haha. Anyway, we went further into derivatives this week and we learned some rules.

The Constant Rule

When you are trying to find a derivative and there is a constant, the derivative of a constant is 0.

The Power Rule

Every time you take a derivative, you lose a power.

So, if you have an exponent, you multiply it by the coefficient and then subtract 1 from the original exponent. Ex. (x^3 would become 3x^2)

Also, we learned how to take derivatives of sine and cosine.

For sin x, the derivative is cos x. For cos x, the derivative is -sin x.

Example:

Find the derivative of (5)/(2x^2).

You would rewrite it as 5/2 x^-2.

Now multiply -2 by 5/2 and you get -5 and raise x^-3 because you subtracted 1. (-5x^-3)

Now just simplify it by writing it as (-5)/(x^3)

The only thing I don't really understand is differentiability. Like where is it differentiable, where is it not.

Blog 3

Chapter two deals with derivatives being differentiable. There are three ways a function is not differentiable. A function is not differentiable if:

1. There is a corner (there is a corner if there is an absolute value or a piecewise)

2. Vertical tangent line

-the exponent is x=^odd/odd

3. If it is not continuous it is either a jump, removable, vertical asymptotes

4. Also, if there is a cusp.

-A cusp is when x=even/odd. When graphing a function a cusp will show up as a dip in the graph.

*Sin x is differentiable, and sin absolute value of x has a corner.

Ex: Where is |x+2| not differentiable?

-Set what is inside equal to 0. X is differentiable at x=-2

Ex: Where is (x+3)^3/7 not differentiable?

-Notice that the exponent is x^odd/odd which makes it a vertical tangent line. The answer would be x=-3

If it asks you to describe the x-values at which f is differentiable you would graph it, and from there remember the rules that make functions differentiable. The answers often are infinity and negative infinity.

Source: Notes:)

Alaina's blog, 12 Sept 2010

This week we learned about the shortcuts to derivatives

and dealing with the derivative of sine and cosine.

First, the shortcut to derivatives is:

1. take the exponent
2. make it the constant
3. reduce the exponent by one
4. plug in

-if it is in a fraction, make it non-fraction by raising x to a negative exponent and then follow the steps.

Ex 1: 2x²+4

=2(2)x¹+4

=4x+4

Ex 2: ²/_3x

=(²/₃)(¹/_x)

=(²>/_3>>)(x^>-1)

=(²/₃>)(-1)(x^-2)

=(-²/₃)(¹>/_x^2sup>)

Second, taking the derivative of sine and cosine

-The rule for sine is

^d/_dx [sin x]=cos x

-The rule for cosine is

^d/_dx [cos x]=-sin x

Ex 1: ^d/_dx [2+2sin x]

=2+2cos x

Ex 2: ^d/_dx[1-cos x]

=1+sin x

Blog #3

For this week we started on chapter two, and it is pretty easy so far. Chapter two deals with derivatives being differentiable. There are three major ways a function is not differentiable. A function is not differentiable if:
1. There is a corner (there is a corner if there is an absolute value or a piecewise)
-Also to see if there is a corner, you graph the function.
2. Vertical tangent line
-A hint for a vertical tangent line is if the exponent is x=^odd/odd
3. If it is not continuous it is either a jump, removable, vertical asymptotes
4. Lastly, if there is a cusp.
-A cusp is when x=even/odd. When graphing a function a cusp will show up as a dip in the graph.
**If a function is differentiable however, it is continuous.
*Sin x is differentiable, and sin absolute value of x has a corner.

Ex: Where is |x+2| not differentiable?
-Set what is inside equal to 0. X is differentiable at x=-2

Ex: Where is (x+3)^3/7 not differentiable?
-Notice that the exponent is x^odd/odd which makes it a vertical tangent line. The answer would be x=-3

**If it asks you to describe the x-values at which f is differentiable you would graph it, and from there remember the rules that make functions differentiable. The answers usually deal with infinity and negative infinity.

-Now for something I don’t understand. I do not understand when the function is a piecewise and from there you have to draw a graph then erase parts of it. I find that really confusing.

Blog #3 - Mary Graci

This is the third Calculus blog of the year. Last week, after learning about taking a derivative, we learned a shortcut. It’s really quite simple, easy to remember, and SO much faster than doing it the long way. Here are the rules:
1. the constant rule: the derivative of a constant is always 0
Ex. 1) d/dx 7 = 0

2. the power rule: whenever you take the derivative of something with an exponent, you lose a power. The formula for this shortcut is d/dx (x^n) = nx^(n-1)
Ex. 2) d/dx 3x^2
First, bring the exponent to the front (and if there is a constant in front of the x, multiply the two). Then, subtract one from the exponent.
= 2(3)x^(2-1) = 6x
Ex. 3) d/dx (x^3 + 9x)
When there are multiple terms in an equation, take the derivative of each term individually.
= 3x^2 + 9

We also learned how to tell if the equation is not differentiable (meaning you can’t take a derivative.) First of all, you it’s not differentiable if there is a corner on the graph. This happens with absolute values, and some piecewise.
Ex. 4) absolute value of (x-3)
There is an absolute value, therefore this equation is not differentiable at x = 3.

Second, you can’t take the derivative of a vertical tangent line (meaning when there is an x raised to an odd # over another odd #).
Ex. 5) (x+5)^3/5
Because of the exponent 3/5, this equation is not differentiable at x = -5.

Third, you can’t take the derivative if it is not continuous (such as with a jump, vertical asymptote, or removable).

And lastly, you can’t take the derivative of a cusp (meaning when there is an x raised to an even # over an odd #).
Ex. 6) x^(6/7)
Because of the exponent 6/7, this equation is not differentiable at x = 0.

I pretty much understood everything this week, especially with the shortcut making everything so much easier!

9/11/10

Alright, so this week the first thing we learned was differentiability (the ability to take a derivative). As in, if a function is not differentiable, then you shouldn't bother trying to take the derivative of it :) To review, a function is not differentiable when there is a "corner" in the graph (which is in absolute value and piecewise functions), when there is a vertical tangent line (when the exponent is an odd number over an odd number), when there is a cusp (graph when exponent is even # over an odd #), and when the function/graph is not continuous (if there's a jump, removable, or vertical asymptote). So, I think that's pretty easy to understand. Also this week we learned the shortcut to finding a derivative!! YAYYYYYYYYYY. And I must say, it is sooo much easier than the long way, however I get confused with some fractions (which I'll say in the "I don't get this stuff" part of this blog)
Anyway, here are a few examples of what I understood the most this week:

Ex. 1) (4x+1)^2 (0,1)
^Find the slope of the graph of the function at the given point.
*The first things you should notice are the key words-"slope of the graph" because that's telling you to take the derivative. Now that you know the shortcut, you can use that here.
*But first, you have to factor the equation. So you "square the first, square the last, 2 times the first times the last"
*So you get: 16x^2 + 8x +1
*Then all you do is take the derivative of each term
*So the derivative of 16x^2 would be 16(2)x = 32x
*d/dx of 8x would be 8
*d/dx of 1 (CONSTANT RULEEEE!) = 0
*So your equation for the slope is 32x + 8
*To get the slope, all you have to do is plug in the x coordinate of the point they gave you.
*So you get 32(0)+8 = 8

Ex. 2) Discuss the differentiability of f(x)=4x+1/x
*Since you may not know what the graph of this looks like, you can type the equation into your y= in your calculator to find out.
*You should see that there are vertical asymptotes (going to infinity and negative infinity) at x=0.
*Therefore, the function is not differentiable because it isn't continuous.
**Hint Hint!!--You really didn't have to type it in your calculator to know there are vertical asymptotes. Since you can't factor the function any more than it already is, and there's still something left in the denominator, setting that equal to zero will give you your vertical asymptotes..

**As for what I didn't understand this week. It's not much actually, I'll just have ALOT of questions to ask on Monday. For instance, I'm having trouble find the derivative of problems like this: (4x^3+3x^2)/(x) ..basically when it's an equation over another equation. I keep working the problems like that over and over, and I can't get the answer they have on calc chat...sooo if anyone can help, that would be appreciated.

Prompt #2--Stephen Ledbetter

Discuss how to find the three main discontinuities and provide an example.

1) jump - A jump exists in a graph when the left and right sides of the limit do not approach the same thing. It is called a jump, because the function jumps from one point to another (limit to limit) and therefore it is discontinuous. This is very prevalent in piecewise functions.

Example:

f(x) =

{x^2 for x is less than 1

{0 for x=1

{2 - (x-1)^2 for x is greater than 1

If you plug in, you will find that there are two different values for 1, 1 and 2. Therefore it is a jump.

2) removable - Removables are mostly seen in factorable functions. It is called removable because you can essentially remove it and the limit could exist continuously after factorization.

Example:

the limit as x approaches 3 of the function

(x^2 - 10x + 21) / (x^2 + x -12)

As all should know, if you plug 3 directly in, you will get an undefined answer. Therefore, you must factor to be able to get a finite limit.

So, that factors out to...

((x-7)(x-3))/((x+4)(x-3))

You can cross out (x-3) on the top and bottom (canceling), therefore you set what you cancel equal to 0 and that is the removable. So x=3 is removable.

3) vertical asymptote - You find this almost the same as a removable. When there is nothing left to factor, you set the bottom equal to 0.

Example:

the limit as x approaches 3 of the function

(x-5)/(x-3)

set x-3=0

x=3 is your vertical asymptote

prompt answer

for some reason when i post as a comment, i start missing pieces of this so....

a.) jumps
a jump is a gap in the graph, its easy to spot and is usually in the form of a piecewise. to see if there is a jump in the graph, plug in for x. if you dont get the same answers than there is a jump. or if there is a fraction you factor as far as possible, cancel what you can, set bottom equal to 0, and solve for x, if the x value you find at the end still causes the bottom to equal 0, there is a jump

lim
x->2
xx^3-4
x
after plugging in 2, we find that the values are not equal, therefore there is a jump

b.)removable
a removable is easy to spot, its a hole in the graph and it always has a limit. to find a removable: factor, cancel, bottom to 0, x=# is the removable

lim
x->2
x3
factor: yes. (x+3)(x-3)/x-3
cancel:x-3 cancels out
x=3 is a removable

c.)vertical asymptote
once again, the x value at the bottom causes it to equal zero after direct substitution. factor, cancel, set bottom to 0, x is the vertical asypmtote

lim x+6/x-5
x->5
factor: already done
cancel: cannot
bottom to zero
x=5 is a vertical asymptote

Prompt #2

For this week’s prompt we have to explain how to find three different types of discontinuities. The three types of discontinuities are a jump, removable, and a vertical asymptote. I guess I will start of by explaining how to know if something is continuous or not. For something to be continuous the left and the right side must be approaching the same number, and if a limit exists, it must be equal to a function value at that point. However, jumps, removables, and vertical asymptotes, are discontinuities and I’ll explain why.

1. Jump
The easiest way to see if a function has a discontinuity is to graph it. When graphing it, if there is a jump in the graph, or a point when the graph will break from one number to another it would be classified as a jump. -How to figure it out without looking at the graph? If the function is x=# and it takes on a different value, it is usually a jump. *A jump could be put in a piecewise function.

2. Removable
There are several ways to find out if a removable has a discontinuity. One way is to try and solve a function. The steps would be: factor and cancel. Take what you cancelled and set it equal to zero. Whatever answer you get (x=#) that would be the removable. In a graph, a removable discontinuity is a point at which a graph is not connected. There will be an open circle at the point where there is a removable.

3. Vertical Asymptote
**Vertical asymptotes occur when it is undefined, infinity, or negative infinity. You can do two things to see if a vertical asymptote has a discontinuity. When graphing the function, the number that does not get touched by the vertical asymptotes will be the discontinuity. When solving without your calculator, follow the same steps as a removable. Except set what is left in the bottom equal to zero. X=# will be the discontinuity.

Week 2 Blog Prompt

Explain how to find each type of discontinuity and provide an example.

a. Jump
b. Removable
c. Vertical Asymptote

Alaina's blog, 5 Sept. 2010

This week, well starting Wednesday, we learned derivatives. They are used to find the slope of a graph or just to find the derivative and are pretty simple.

Notation for Derivative:

-Slope of a tangent line equals a derivative

1. ^dy/_dx
2. y¹
3. f¹(x)
4. ^d/_dx[f(x)]
5. Dx[y]

Basic Formulas:

Slope of a secant line: ^{f(x+Δx)-f(x)}/_Δx

Slope of a tangent line: _lim^{f(x+Δx)-f(x)/_Δx}

^2->0

Ex 1: find the slope of th egraph of f(x)=2x-1 at (2,1)

_{lim =}^{2(x+Δx)-1(2x-1)}/_Δx

^x->0

=^{2x+2Δx-1-2x-1}/_Δx

=^2Δx/_Δx

=Δx

Ex 2: Find the derivative of f(x)=x²+4x

_{lim =} (x+Δx>²+4(x+Δx)-(x²+4x)/_Δx

^x->0

=x²+2Δxx+Δx²+4x+4Δx-x²-4x/_Δx

=4Δx

I'm not really sure whether or not I did the last one correctly, but I think that I did. If I did not, feel free to correct my work.

2nd Blog

This week we learned derivatives. Its basically just plugging in and simple algebra. The main formula used is the slope of a secant line:

(f(x+dx)-f(x))/(dx)

But, sometimes you have to use another formula, which is for the slope of a tangent line. It is the same formula, except that it is a limit approaching zero in the tangent line version.

Ex. f(x)=x^2+x-3

We just plug in to the tangent slope formula.

((x+dx)^2+(x+dx)-3-(x^2+x-3))/(dx)

(x^2+2xdx+dx^2+x+dx-3-x^2-x+3)/(dx)

(2xdx+dx^2+dx)/(dx)

(dx(2x+dx+1))/(dx)

2x+dx+1

2x+1

Blog 2

Well, looks like this week was filled with math........how "fun" was that.......
I really didn't get it too much towards the end of the week.
So I shall talk of dragons, bards, music, and fanciful things........jk, can't be funny on here because it's "not appropriate".......we shall see random person....we shall see...... :P

Anyways, instead of all of that epic stuff from before, I'll just talk about limits, because I didn't really understand the stuff from the end of the week.

let's say you wanna find the limit of an equation, such as lim ((x-3)^2)/x
x->7

You've got to plug in whatever x approaches into the equation, as long as it doesn't result in a 0 on the bottom of the equation.

So if it were x -> 0 instead of x -> 7, the answer to this problem would most likely be Ø.
If that happens, you need to do one of two things.
*note* You do this whenever the bottom would equal Ø when you plug in whatever x approaches, not just when x -> 0.

One way would be by plugging the function into a table and using numbers that are close to the x -> to find what both sides approach.
In this case, the table would be: -.1 -.01 -.001 0 .001 .01 .1
-96.1 -906 -9006 8994 894.01 84.1
So as you can see, from the left, the graph/equation is approaching -∞, and from the right, the graph/equation is approaching ∞.

Another way you could figure it out is by using arithmetic.
((x-3)^2)/x
(x^2-6x+9)/x = 0
x^2-6x+9=0
x^2-6x=-9
x(x-6)=-9
x=-9 x=6
And then you figure out which is the limit, which it usually is hard to figure out.

Now, I'm getting the fudge out of here before all these monkeyfudgers get all this math into my fudging head.
FUDGE!! I strongly dislike math......as you can see already...... :P
Now, I'm gonna go fudge up some fires.

*if you don't get the whole fudging thing, check out this vid, it explains a lot, and it's funny XD*
http://www.youtube.com/watch?v=uovMpapeCJQ

Blog 2

Soooo we learned about derivatives and stuff in class this week. There are only two formulas you need to know so it aint that hard.

the two formulas are

f(x+dx)-f(x)/(dx) - secant line

lim as x -> 0 f(x+dx)-f(x)/(dx) - tangent line

the slope of a tangent line is the derivative of something.

Example:

Find the slope of the graph of f(x) = x^2 + 3x + 1 at (1, 3)

lim as Dx approaches 0: ( f(x + Dx) - f(x) ) / (Dx)

What you have to do is plug in (x+ Dx) for every x in the formula.

(x + Dx)^2 + 3(x + Dx) + 1 - (x^2 + 3x + 1) / (Dx)

That simplifies to...

x^2 + 2xDx + Dx^2 + 3x + 3Dx + 1 - x^2 - 3x -1 / Dx

2xDx + Dx^2 + 3Dx / Dx

2x + Dx + 3

plug in 0 for Dx

2x + 3 is your equation

Now just plug in the x coordinate to find the slope.

2(1) + 3 = 5

The slope is 5.

wellll i dont know how many words this is supposed to be so someone let me know. have a fun monday people.

2nd blog

this week we started derivatives, which are pretty easy to do
the formulas:
f(x+dx)-f(x) is for slope of a secant line (d=delta)
-----------
dx

lim f(x+dx)-f(x)
dx->0 ------------ is slope of tangent line
dx

the slope of tangent line formula is used to find a derivative, tests can try to trick you by asking several different names that mean the same
first plug in the function given to the formula, after finding the solution to the formula, plug in the point given to find the slope of the equation
to see if its correct you check on the calculator

Blog #2 - Mary Graci

This is the second blog of Calculus! Anyway, at the beginning of the week, we reviewed for the Chapter 1 test we had Wednesday. Thursday, we started on derivatives. For derivatives, you use the slope of a tangent line formula. This formula is: lim as x approaches 0 f(x+delta x) - f(x) / delta x

When you are asked to find the derivative, it can be asked in a variety of ways: f'(x), dx/dy, etc.

Ex. 3x
First, plug it into the formula. 3(x+delta x) - 3x / delta x
Then, distribute and simplify. 3x + 3delta x - 3x / delta x
Cancel out whatever you can. 3delta x / delta x
Cancel again to get the delta x out of the denominator, and solve. = 3

Ex. x^2
Again, plug into the formula. (x+delta x)^2 - x^2 / delta x
Distribute and simplify. (x^2 + 2x delta x + delta x^2) - x^2 / delta x
Cancel. 2x delta x + delta x^2 / delta x
Factor out a delta x, then cancel again. delta x(2x + delta x) / delta x = 2x + delta x
Plug in 0 for delta x. 2x + (0) = 2x

Next week, we'll learn a shortcut of how to find a derivative. As for this week, I understood just about everything and I even did really good on my first Calculus test! Anyway, I don't have any questions for this week. Good luck on the upcoming weeks in Calc, everyone!

Blog 2

Alright, so this is the second blog and like the third or fourth week in calculus. I’m doing better than I thought I would be doing. The class is not too hard, you just really have to pay attention and do the homework. And I really like that we get a huge homework packet before each test, it really helps me a lot. Anyways, we just started chapter two which is not too hard but some of the problems can get complicated and confusing at times. Chapter two deals with two formulas: f(x+Deltax) – f(x)/deltax (Slope of secant line) and the second formula, lim as deltax approaches 0 f(x + deltax_ - f(x)\deltax (Slope of tangent line). These two formulas are the exact same expect that the slope of the tangent line has the limit as deltax approaches 0. **Also every time it asks to find the derivative, the second formula will be used. To use the formulas, the problem will give you a function and you will just have to plug it into the formula. Then it will give you a point and once you figure out the whole problem then you will find the slope with the EQUATION that is left. After finding slope you plug into the point slope formula to get the answer.

Ex: h(t)=t^2+3, (-2,7)
(x+deltax)^2+3-(t^2+3)
X^2+2xdeltax+deltax^2-t^2-3/delta x
(factor out a deltax and it will cancel)
2(-2)=-4

**The thing that I don’t understand about this chapter is what to do when a function is like x^3. One example is problem 35 on page 104.

9/5/10

Okay so this week we had our Chapter 1 limits test..prettyyyyy easy if I do say so myself. Then on Thursday we started derivatives, the long way, which seems easy so far so that's good. We learned the key words for derivatives, such as: dy/dx, y', f'(x), [f(x)], Dx[y]. **And a derivative is also defined as the slope of a tangent line. So I'd say it's pretty important to know these things so you'll be sure to know what a problem is asking you to find. Anyway, I thought pretty much everything we went over this week was easy. So here's a few examples of what we did:

Ex 1) Find the slope of the tangent line of the function f(x)=x^2+3x+4
**First off you have to figure out what the problem is asking you to find. "The slope of the tangent line" is a....DERIVATIVE! So all you have to do for this problem is find the derivative of the function given to you.
*So in order to find the derivative, all you do is plug the function into the derivative formula:
f(x+#x)-f(x)/#x *where lim#x>0...Which means all you do is plug (x+#x) into the equation every time you see an "x" ...(**# will be delta) and then subtract the original equation after. Then simplify the equation, and plug in zero as #x after.
*So you get: (x+#x)^2+3(x+#x)+4-(x^2+3x+4)
*Then expanding that you get:
x^2+2x#x+#x^2+3x+3#x+4-x^2-3x-4 / #x
(*Don't forget to distribute the negative to the original equation)
*Then you can cancel out x^2 and -x^2, 3x and -3x, 4 and -4
*So you have 2x#x+#x^2+3#x/#x
*Now you can factor out #x, since that's the only like term. And that'll cancel out the #x on the denominator also
*So after factoring #x out you get: #x(2x+#x+3)/#x
*Then you can cancel out the #x on the numerator and denominator and plug zero into your final equation >> 2x+#x+3
*So your slope of the tangent line is 2x+3

Ex. 2) Find an equation of the tangent line to the graph of f(x) at the given point.
f(x)=x^3 (2,8)
*The key word of this problem is "tangent line" which tells you you're going to have to find the derivative first
*So you get (x+#x)^3
(x+#x)(x^2+2x#x+#x^2)-x^3 /#x
x^3 + 2x#x + x#x^2 + x^2#x + 2x#x^2 + #x^3 - x^3 / #x
*Now you can cancel out x^3 and -x^3
*So now you have 2x^2#x + x#x^2 + x^2#x + 2x#x^2 + #x^3 / #x
*Next you can factor out a #x
#x(2x^2 + x#x + x^2 2x#x + #x^2) / #x
*then the #x on the numerator and denominator cancel out, and you can plug zero into the equation as #x
*So simplifying that equation you get 3x^2
*Now that you've got the derivative, you have to use the point they gave you, (2,8) to get the slope.
*So to get the slope of 2 (2 being the x term), you plug 2 into 3x^2.
3(2)^2 = 12
*Now to get your final equation you plug the slope and the point into the point slope formula:
y-y1=m(x-x1)
**So you're equation is y-8=12(x-2)

*I basically understood everything we learned this week. The only thing I'm unsure of is how to "discuss the behavior of f at (0,0)" ..like example 4 from our notes.

Blog #2 - Stephen Ledbetter

This week was pretty good actually. I'm so glad we have a 3-day weekend haha. I found out in the lesson this week that derivatives and slope stuff is really easy. There are two formulas that we learned this week. Slope of secant line and slope of tangent line formulas (also known as the derivative formula). And the two formulas are:

(D = delta)

slope of secant line formula: ( f(x + Dx) - f(x) ) / (Dx)

slope of tangent line formula (derivative): lim as Dx approaches 0: ( f(x + Dx) - f(x) ) / (Dx)

The only difference is that for the derivative formula you plug in 0 for Dx towards the end of the "solving" process.

Example:

Find the slope of the graph of f(x) = x^2 + 3x + 1 at (1, 3)

lim as Dx approaches 0: ( f(x + Dx) - f(x) ) / (Dx)

What you have to do is plug in (x+ Dx) for every x in the formula.

So...

(x + Dx)^2 + 3(x + Dx) + 1 - (x^2 + 3x + 1) / (Dx)

That simplifies to...

x^2 + 2xDx + Dx^2 + 3x + 3Dx + 1 - x^2 - 3x -1 / Dx

2xDx + Dx^2 + 3Dx / Dx

2x + Dx + 3

plug in 0 for Dx

2x + 3 is your equation

Now just plug in the x coordinate to find the slope.

2(1) + 3 = 5

The slope is 5.

There's not much I don't really understand except that I'm having a little trouble with the derivatives with the square roots in them. I'm not good with conjugates.

Taylor Rodriguez : Blog 2

This week we covered one major point of chapter two. This is the point that I will summarize in this week’s blog.
There are two formulas which need to be memorized
(((& means delta)
• The first formula is
f(x+&x)- f(x)/&x
This is the formula for a derivative. This formula is known as the secant line formula.

• The second formula is only a tiny bit different from the first

Lim f(x+&x)- f(x)/&x
&x -> 0

This formula is known as the slope of a tangent line

Solving for the problems we’ve had thus far in chapter 2 have consisted of plugging into these formulas and solving.

When given an equation you must plug x+&x into all x’s of the given equation and fill in the rest of the formula by placing – f(given equation exactly how its given)/ &x

Ex:
Find the slope of the graph of f(x)= 2x-3 at (2,1)

First you would plug in:

f(x) 2(x+&x)-3-(2x-3)/ &x
&x-> 0

Expand:

2x+2&x – 3 – 2x + 3 / &x

Take out what cancels:
2x + 2&x -3 -2x +3/&x

And you’re left with:
2&x/&x

Simplify:

2 &x/ &x
Therefore the answer is 2

((because there are no x’s left in the equation you can ignore the point (2,1) however if there had been any x’s you would have also plugged in 2 for the x’s before you simplified then solved as normal.))

There are a few helpful hints that need to be remembered
**remember that for each of these formulas any variable can be used for delta x
** remember that slope of the tangent line means to use the derivative formula
** there are many was to ask for a derivative these ways are:
Dy/dx
Y^1
F’(x)
d/dx {f(x)}
Dx[y]
D/dx

Prompt 1

For this week’s prompt we have to explain three statements. It would be helpful to know if a jump, removable, or asymptote could even be continuous. Knowing if something is continuous or not would help you know if it could have a limit. To know if something is continuous, draw a graph from start to finish without picking up the pencil. **Therefore, a function is continuous if the graph has no holes or breaks in it. Now to explain the three statements.

1.A jump NEVER has a limit
-A jump will never have a limit because a jump will always have a break when you graph the function. When there is a jump the values will not match up because of the break it the graph. A jump also makes the function not continuous because there is no point, and the graph has to stop. It would be hard to tell if a problem was a jump by just looking at it, you would have to plug it into your calculator and show the graph to figure it out.

2.An asymptote (infinite) might have a limit
-There are several ways to figure out if an asymptote has a limit or not. One thing to remember is an asymptote is always a fraction (the function). The first way to figure out if a function is an asymptote is it out by trying to factor it. After factoring, if nothing cancels at the bottom then you set it equal to 0. X= the number would be the asymptote. Another way to figure it out is to plug the function into your calculator, then to graph it. If the graph has two curves that do not intersect, and they approach infinity or negative infinity, the function will have an asymptote. The limit of the asymptote will be the point between the two curves. The point will never touch the curves.

3.A removable ALWAYS has a limit
-If you are looking at a graph, a removable usually has a hole where the limit is. However, just because the hole shows that it is a discontinuity, does not mean it isn’t a limit. **Note that also a removable will be a fraction. When trying to figure out whether or not a problem is a removable, and has a limit, without using your calculator you would follow the same steps as an asymptote. First, you would factor what you could. Next, you would cancel out the things you could from the top and bottom. This step is different from an asymptote, set what you cancel equal to 0 and whatever x= would be the limit.