Derivatives Part A. Review of Basic Rules f(x)=xf`(x)=1 f(x)=kx f`(x)= k f(x)=kx n f`(x)= (k*n)x...

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Derivative s Part

Transcript of Derivatives Part A. Review of Basic Rules f(x)=xf`(x)=1 f(x)=kx f`(x)= k f(x)=kx n f`(x)= (k*n)x...

Page 1: Derivatives Part A. Review of Basic Rules f(x)=xf`(x)=1 f(x)=kx f`(x)= k f(x)=kx n f`(x)= (k*n)x (n-1)    1.) The derivative of a variable is 1. 2.)

Derivatives

Part A

Page 2: Derivatives Part A. Review of Basic Rules f(x)=xf`(x)=1 f(x)=kx f`(x)= k f(x)=kx n f`(x)= (k*n)x (n-1)    1.) The derivative of a variable is 1. 2.)

Review of Basic Rules

f(x)=x f`(x)=1

f(x)=kx f`(x)= k

f(x)=kxn f`(x)= (k*n)x (n-1)

1.) The derivative of a variable is 1.

2.) The derivative of a constant multiplied by a variable is the constant.

3.) The derivative of a variable to a power is found by multiplying the coefficient by the power and then subtracting 1 from the power.

Page 3: Derivatives Part A. Review of Basic Rules f(x)=xf`(x)=1 f(x)=kx f`(x)= k f(x)=kx n f`(x)= (k*n)x (n-1)    1.) The derivative of a variable is 1. 2.)

Review of Basic Rules

f(x)=sin(X) f`(x)=cos(X)

f(x)=g(x) ± h(x) f`(x)= g`(x) ± h`(x)

f(x)=cos(X) f`(x)= - sin(X)

4.) The derivative of the trig function Sine is Cosine.

5.) The derivative of the trig function Cosine is Sine.

6.) If you have terms that are added or subtracted you may find the derivative of each term separately and the combine the derivative.

Page 4: Derivatives Part A. Review of Basic Rules f(x)=xf`(x)=1 f(x)=kx f`(x)= k f(x)=kx n f`(x)= (k*n)x (n-1)    1.) The derivative of a variable is 1. 2.)

The Product Rule : When you have expressions multiplied , you assign the temporary variable “u” to one expression and the temporary variable “v” to the other. Then you find the derivative of “U” and call it “du”. Find the derivative of “V” and call it “dv”. Insert the values into the product rule and evaluate it.

Page 5: Derivatives Part A. Review of Basic Rules f(x)=xf`(x)=1 f(x)=kx f`(x)= k f(x)=kx n f`(x)= (k*n)x (n-1)    1.) The derivative of a variable is 1. 2.)

Product Rule Review:

f(x) = 3x2 ( x3 + 2 )

f `(x) = uv` + vu`

f `(x) = ( )( ) + ( )( )

u v

v` = 3x2

u` = 6x

3x2 x3 + 2

9x4 + 6x4 + 12x

15x4 + 12x

3x2

6x

Page 6: Derivatives Part A. Review of Basic Rules f(x)=xf`(x)=1 f(x)=kx f`(x)= k f(x)=kx n f`(x)= (k*n)x (n-1)    1.) The derivative of a variable is 1. 2.)

The Quotient Rule : When you have expressions divided , you assign the temporary variable “u” to the numerator’s expression and the temporary variable “v” to the denominator’s expression. Then you find the derivative of “U” and call it “du”. Find the derivative of “V” and call it “dv”. Insert the values into the quotient rule and evaluate it.

Page 7: Derivatives Part A. Review of Basic Rules f(x)=xf`(x)=1 f(x)=kx f`(x)= k f(x)=kx n f`(x)= (k*n)x (n-1)    1.) The derivative of a variable is 1. 2.)

Quotient Rule:

f(x) = x2 + 3

x + 5

f`(x) = vu`- uv`

v2

( )( ) – ( )( )

( )2

2x v` = 1

u` =

f `(x) =

2x2 + 10x

x2 +10x + 25

x2 + 10x - 3

u

v x2 + 3 2x

1x + 5

x2 3- -x2 +10x + 25

=

f `(x) =

x + 5

- -

Page 8: Derivatives Part A. Review of Basic Rules f(x)=xf`(x)=1 f(x)=kx f`(x)= k f(x)=kx n f`(x)= (k*n)x (n-1)    1.) The derivative of a variable is 1. 2.)

Notation:f(x) This notation tells the user that the answer depends on how the variable x changes. We say that the answer is a “Function” of x.

dx

dydx

du

This notation indicates we are getting the 1st derivative of y based on x.

du

dy

Means: Find the 1st derivative of y based on the variable u

Means: Find the 1st derivative of u based on the variable x

Page 9: Derivatives Part A. Review of Basic Rules f(x)=xf`(x)=1 f(x)=kx f`(x)= k f(x)=kx n f`(x)= (k*n)x (n-1)    1.) The derivative of a variable is 1. 2.)

The

Chain Rule

Page 10: Derivatives Part A. Review of Basic Rules f(x)=xf`(x)=1 f(x)=kx f`(x)= k f(x)=kx n f`(x)= (k*n)x (n-1)    1.) The derivative of a variable is 1. 2.)

The Chain Rule states that if you have a composite function, a function within a function, you can break down the function to simplify it.

dx

du

du

dy

dx

dy

Many times we can rewrite a complicated function as a composite, which will allow us to use our basic derivative rules to find the derivative.

Page 11: Derivatives Part A. Review of Basic Rules f(x)=xf`(x)=1 f(x)=kx f`(x)= k f(x)=kx n f`(x)= (k*n)x (n-1)    1.) The derivative of a variable is 1. 2.)

Examples of breaking down a composite function.

Page 12: Derivatives Part A. Review of Basic Rules f(x)=xf`(x)=1 f(x)=kx f`(x)= k f(x)=kx n f`(x)= (k*n)x (n-1)    1.) The derivative of a variable is 1. 2.)

Examples of breaking down a composite function.

Page 13: Derivatives Part A. Review of Basic Rules f(x)=xf`(x)=1 f(x)=kx f`(x)= k f(x)=kx n f`(x)= (k*n)x (n-1)    1.) The derivative of a variable is 1. 2.)

Examples of breaking down a composite function.

Page 14: Derivatives Part A. Review of Basic Rules f(x)=xf`(x)=1 f(x)=kx f`(x)= k f(x)=kx n f`(x)= (k*n)x (n-1)    1.) The derivative of a variable is 1. 2.)

Steps for using the CHAIN RULE

Page 15: Derivatives Part A. Review of Basic Rules f(x)=xf`(x)=1 f(x)=kx f`(x)= k f(x)=kx n f`(x)= (k*n)x (n-1)    1.) The derivative of a variable is 1. 2.)

Examples of using the CHAIN RULE

Page 16: Derivatives Part A. Review of Basic Rules f(x)=xf`(x)=1 f(x)=kx f`(x)= k f(x)=kx n f`(x)= (k*n)x (n-1)    1.) The derivative of a variable is 1. 2.)

Examples of using the CHAIN RULE

Page 17: Derivatives Part A. Review of Basic Rules f(x)=xf`(x)=1 f(x)=kx f`(x)= k f(x)=kx n f`(x)= (k*n)x (n-1)    1.) The derivative of a variable is 1. 2.)

Examples of using the CHAIN RULE

Page 18: Derivatives Part A. Review of Basic Rules f(x)=xf`(x)=1 f(x)=kx f`(x)= k f(x)=kx n f`(x)= (k*n)x (n-1)    1.) The derivative of a variable is 1. 2.)

Examples of using the CHAIN RULE

Page 19: Derivatives Part A. Review of Basic Rules f(x)=xf`(x)=1 f(x)=kx f`(x)= k f(x)=kx n f`(x)= (k*n)x (n-1)    1.) The derivative of a variable is 1. 2.)

Examples of using the CHAIN RULE