since this week we mainly focused on the uno study i am going to review again before we start doing the ap review tests.
I want to give a review on all things derivative because to me majority of the calculus we have learned has some form of derivative involved.
SLOPE OF A TANGENT LINE
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
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
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.
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
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
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