Made by:

Rudra Patel

Harsh Desai

Ravi Patel

Rishabh Patel

Harshil Raymagiya

We now have a pretty good list of “shortcuts” to find derivatives of simple functions.

Of course, many of the functions that we will encounter are not so simple. What is needed is a way to combine derivative rules to evaluate more complicated functions.

→

Consider a simple composite function:

6 10y x= −

( )2 3 5y x= −

If 3 5u x= −

then 2y u=

6 10y x= − 2y u= 3 5u x= −

6dy

dx= 2

dy

du= 3

du

dx=

dy dy du

dx du dx= ⋅

6 2 3= ⋅

→

and another:

5 2y u= −

where 3u t=

( )then 5 3 2y t= −

3u t=

15dy

dt= 5

dy

du= 3

du

dt=

dy dy du

dt du dt= ⋅

15 5 3= ⋅

( )5 3 2y t= −

15 2y t= −

5 2y u= −

→

and one more:29 6 1y x x= + +

( ) 23 1y x= +

If 3 1u x= +

3 1u x= +

18 6dy

xdx

= + 2dy

udu

= 3du

dx=

dy dy du

dx du dx= ⋅

2y u=

2then y u=

29 6 1y x x= + +

( )2 3 1dy

xdu

= +

6 2dy

xdu

= +

( )18 6 6 2 3x x+ = + ⋅This pattern is called the chain rule.

→

dy dy du

dx du dx= ⋅Chain Rule:

If is the composite of and , then:f go ( )y f u= ( )u g x=

( ) ( ) at at xu g xf g f g=′ ′ ′= ⋅o

example: ( ) sinf x x= ( ) 2 4g x x= − Find: ( ) at 2f g x′ =o

( ) cosf x x′ = ( ) 2g x x′ = ( )2 4 4 0g = − =

( ) ( )0 2f g′ ′⋅

( ) ( )cos 0 2 2⋅ ⋅

1 4⋅ 4=→

We could also do it this way:

( )( ) ( )2sin 4f g x x= −

( )2sin 4y x= −

siny u= 2 4u x= −

cosdy

udu

= 2du

xdx

=

dy dy du

dx du dx= ⋅

cos 2dy

u xdx

= ⋅

( )2cos 4 2dy

x xdx

= − ⋅

( )2cos 2 4 2 2dy

dx= − ⋅ ⋅

( )cos 0 4dy

dx= ⋅

4dy

dx=

→

Here is a faster way to find the derivative:

( )2sin 4y x= −

( ) ( )2 2cos 4 4d

y x xdx

′ = − × −

( )2cos 4 2y x x′ = − ×

Differentiate the outside function...

…then the inside function

At 2, 4x y′= =

→

Another example:

( )2cos 3d

xdx

( ) 2cos 3

dx

dx

( ) ( )2 cos 3 cos 3d

x xdx

⋅

derivative of theoutside function

derivative of theinside function

It looks like we need to use the chain rule again!

→

Another example:

( )2cos 3d

xdx

( ) 2cos 3

dx

dx

( ) ( )2 cos 3 cos 3d

x xdx

⋅

( ) ( ) ( )2cos 3 sin 3 3d

x x xdx

⋅ − ⋅

( ) ( )2cos 3 sin 3 3x x− ⋅ ⋅

( ) ( )6cos 3 sin 3x x−

The chain rule can be used more than once.

(That’s what makes the “chain” in the “chain rule”!)

→

Derivative formulas include the chain rule!

1n nd duu nu

dx dx−= sin cos

d duu u

dx dx=

cos sind du

u udx dx

= − 2tan secd du

u udx dx

=

etcetera…

The formulas on the memorization sheet are written with

instead of . Don’t forget to include the term!

u′u′du

dx→

The most common mistake on the chapter 3 test is to forget to use the chain rule.

Every derivative problem could be thought of as a chain-rule problem:

2dx

dx2d

x xdx

= 2 1x= ⋅ 2x=

derivative of outside function

derivative of inside function

The derivative of x is one.

→

The chain rule enables us to find the slope of parametrically defined curves:

dy dy dx

dt dx dt= ⋅

dydydt

dx dxdt

=

Divide both sides bydx

dtThe slope of a parametrized curve is given by:

dydy dt

dxdxdt

=

→

These are the equations for an ellipse.

Example: 3cosx t= 2siny t=

3sindx

tdt

= − 2cosdy

tdt

= 2cos

3sin

dy t

dx t=

−2

cot3

t= −

π