DERIVATIVESBy: Susana Cardona & Demetri

Cheatham

© Cardona & Cheatham 2011

DIFFERENTIATION A.K.A DERIVATIVE

Slope of a tangent line

Six different techniques: Chain rule, product rule, Quotient rule, E.T.A, Implicit differentiation and Logs.

Chain Rule

Bring exponent down in front of the variable, if it’s a coefficient multiply exponent. Then subtract one from the exponent and go back in and take a derivative.

1

( )

( )

n

n

f x ax

f x anx

Example

3 4

2 3

( ) 6 4

( ) 18 16

f x x x

f x x x

3

2

( ) (5 1)

( ) 3(5 1) (5)

f x x

f x x

Try Me 3 9( ) ( 7 )f x x x

Solution

3 8 2( ) 9( 7 ) (3 7)f x x x x

Product Rule

First write the problem times derivative of the second problem plus write the second problem times the derivative of the first problem.

FDS+SDF

Example

2 3 3 7(5 1) (2 4)x x 2 3 3 6 2 3 7 2 4(5 1) (7)(2 4) (6 ) (2 4) ( 3)(5 1) (10 )x x x x x x

2 4( ) ( 3) (3 1)f x x x 2 3 4( ) ( 3) (4)(3 1) (3) (3 1) (2)( 3)(1)f x x x x x

( )f x

( )f x

Try Me

4 3(3 1) (1 2 )x x ( )f x

Solution

4 4 3 3(3 1) ( 3)(1 2 ) ( 2) (1 2 ) (4)(3 1) (3)x x x x ( )f x

Quotient Rule

Write the bottom times the derivative of the top minus write the top times the derivative of the bottom over the bottom squared

2

BDT TBD

B

Example

3

4

(5 1)

(2 1)

x

x

4 2 3 3

8

(2 1) (3)(5 1) (5) (5 1) (4)(2 1) (2)

(2 1)

x x x x

x

( )f x

( )f x

Try Me

◦ 2 2

2 4

(3 5 )

(6 2 )

x x

x x

( )f x

Solution

2 4 2 2 2 2 4 3

2 4 2

(6 2 )(2)(3 5 )(6 5) (3 5 ) (6 2 )(12 8 )

(6 2 )

x x x x x x x x x x x

x x

( )f x

ETA A.K.A Exponent, Trig, Angle

Bring down exponent, multiply coefficient if there’s one, and write the trig and the angle times the derivative of the trig times the derivative of the angle

Example

1.

2.

2(sin 3 )d

xdx

3(cos (sin ))d

xdx

23cos (sin ) ( sin(sin )) (cos )x x x

co2s s33 3in x

E

x

T A

Try Me

3xde

dx

Solution

3 (3)xe

NATURAL LOG

1 over the angle times the derivative of the angle

ln

x

x

y a

dya a

dx

EXAMPLE

1(0) ln 2(1)

dyx

y dx

1ln 2

dy

y dx

2 ln 2xdy

dx

2xy ln ln 2y x

TRY ME

ln xy x

SOLUTION

ln

ln ln ln

1 1 1ln ( ) ln ( )

2 ln( )x

y x x

dyx x

y dx x x

dy xx

dx x

Implicit

•Is almost the same as a chain rule but it includes x and y and the x’s and y’s can be separated

•

2 2 1

2 2 0

2

2

x y

dyx y

dxdy x

dx y

dy x

dx y

Example3 2

2

2

2

3 4 5

9 8 0

8 9

9

8

x y

dyx y

dxdyy xdx

dy x

dx y

•

Try Me2 23 2 5 1x xy y

Solution

6 2 (1) 2 (1) 10 0

( 2 10 ) 2 6

2 6

2 10

3

5

dy dyx x y y

dx dxdy

x y y xdxdy y x

dx x y

dy y x

dx x y

Practice Problem

4 2(tan ( 2 1))d

x xdx

Solution

3 2 2 24 tan ( 2 1) sec ( 2 1) (2 2)x x x x x

PRACTICE PROBLEM

22 1y x x

OR

12 2(2 )( 1)y x x

SOLUTION

1 1

2 22 21

' (2 )( )( 1) (2 ) ( 1) (2)2

y x x x x

Practice Problem

3

2

(5 1)y

x

Solution

4' 6(5 1) (5)y x