MAT01A1: the Chain Rule

Dr Craig

Week: 4 May 2020

This lecture covers a very important

differentiation technique. Before we get

started, let’s recap some of the concepts

from last week.

Warm up:

A nice question that requires Special Limits:

limθ→0

sin 3θ

θ + tan 8θAttempt this and then look at the solution

on the next slide.

Hint: look at the example limx→0

sin 7x

4xfrom

last time.

Solution: We want to get this in a form so

that we can use limx→0

sinx

x= 1. We divide

above and below by 3θ.

limθ→0

sin 3θ

θ + tan 8θ= lim

θ→0

(sin 3θ)/(3θ)

θ/(3θ) + (tan 8θ)/(3θ)

Next, we change the tan 8θ tosin 8θ

cos 8θ.

limθ→0

sin 3θ3θ

13 +

tan 8θ3θ

= limθ→0

sin 3θ3θ

13 +

sin 8θ3θ cos 8θ

Solution: We want to get this in a form so

that we can use limx→0

sinx

x= 1. We divide

above and below by 3θ.

limθ→0

sin 3θ

θ + tan 8θ= lim

θ→0

(sin 3θ)/(3θ)

θ/(3θ) + (tan 8θ)/(3θ)

Next, we change the tan 8θ tosin 8θ

cos 8θ.

limθ→0

sin 3θ3θ

13 +

tan 8θ3θ

= limθ→0

sin 3θ3θ

13 +

sin 8θ3θ cos 8θ

Solution: We want to get this in a form so

that we can use limx→0

sinx

x= 1. We divide

above and below by 3θ.

limθ→0

sin 3θ

θ + tan 8θ= lim

θ→0

(sin 3θ)/(3θ)

θ/(3θ) + (tan 8θ)/(3θ)

Next, we change the tan 8θ tosin 8θ

cos 8θ.

limθ→0

sin 3θ3θ

13 +

tan 8θ3θ

= limθ→0

sin 3θ3θ

13 +

sin 8θ3θ cos 8θ

Solution: We want to get this in a form so

that we can use limx→0

sinx

x= 1. We divide

above and below by 3θ.

limθ→0

sin 3θ

θ + tan 8θ= lim

θ→0

(sin 3θ)/(3θ)

θ/(3θ) + (tan 8θ)/(3θ)

Next, we change the tan 8θ tosin 8θ

cos 8θ.

limθ→0

sin 3θ3θ

13 +

tan 8θ3θ

= limθ→0

sin 3θ3θ

13 +

sin 8θ3θ cos 8θ

Now we apply the same technique that we

used for the example in the last lecture:

limθ→0

sin 3θ3θ

13 +

sin 8θ3θ cos 8θ

= limθ→0

sin 3θ3θ

13 +(

sin 8θ3θ cos 8θ ×

88

)

We re-arrange the fraction in the bottom

right and take the limit:

limθ→0

sin 3θ3θ

13 +(

83 cos 8θ ×

sin 8θ8θ

) = limθ→0

sin 3θ3θ

limθ→0

13 + lim

θ→0

83 cos 8θ limθ→0

sin 8θ8θ

=1

(1/3) + (8/3)(1)=

1

3

Now we apply the same technique that we

used for the example in the last lecture:

limθ→0

sin 3θ3θ

13 +

sin 8θ3θ cos 8θ

= limθ→0

sin 3θ3θ

13 +(

sin 8θ3θ cos 8θ ×

88

)We re-arrange the fraction in the bottom

right and take the limit:

limθ→0

sin 3θ3θ

13 +(

83 cos 8θ ×

sin 8θ8θ

) = limθ→0

sin 3θ3θ

limθ→0

13 + lim

θ→0

83 cos 8θ limθ→0

sin 8θ8θ

=1

(1/3) + (8/3)(1)=

1

3

Now we apply the same technique that we

used for the example in the last lecture:

limθ→0

sin 3θ3θ

13 +

sin 8θ3θ cos 8θ

= limθ→0

sin 3θ3θ

13 +(

sin 8θ3θ cos 8θ ×

88

)We re-arrange the fraction in the bottom

right and take the limit:

limθ→0

sin 3θ3θ

13 +(

83 cos 8θ ×

sin 8θ8θ

) = limθ→0

sin 3θ3θ

limθ→0

13 + lim

θ→0

83 cos 8θ limθ→0

sin 8θ8θ

=1

(1/3) + (8/3)(1)=

1

3

Quick revision quiz

Write down the following

1. (fg)′

2.

(f

g

)′3. d

dx(sinx)

4. ddx(cosx)

5. ddx(tanx)

6. ddx(cscx)

7. ddx(secx)

8. ddx(cotx)

The product rule:

If f and g are both differentiable, then:

d

dx[f (x).g(x)] = f (x)

d

dxg(x)+g(x)

d

dxf (x)

Or, in abbreviated form:

(f · g)′ = f ′g + g′f

The quotient rule:

If f and g are both differentiable, then:

d

dx

[f (x)

g(x)

]=g(x) ddx[f (x)]− f (x)

ddx[g(x)]

[g(x)]2

In abbreviated form:(fg

)′=f ′g − g′f

g2.

Derivatives of Trig Functions:

d

dx(sinx) = cosx

d

dx(cscx) = − cscx cotx

d

dx(cosx) = − sinx

d

dx(secx) = secx tanx

d

dx(tanx) = sec2 x

d

dx(cotx) = − csc2 x

I regard the Chain Rule as the single mostimportant topic in the entire year. If you

do not properly understand the Chain Rule,

the rest of MAT01A1 and MAT01B1 will be

a total mystery. So, here we go. . .

Derivatives of more complex functions

Our rules do not yet allow us to differentiate

functions like the following

F (x) =√x2 + 1

G(x) = sin(3x)

H(x) = ex2

The Chain Rule enables us to differentiate

composite functions. A proof can be found

in the textbook (it is not required).

The Chain Rule

If f and g are both differentiable and

F = f ◦ g is the composite function

F (x) = f (g(x)),

then F is differentiable and F ′ is given by

F ′(x) = f ′(g(x)).g′(x)

Example: Findd

dx

(√x2 + 3x

)

The Chain Rule

If f and g are both differentiable and

F = f ◦ g is the composite function

F (x) = f (g(x)),

then F is differentiable and F ′ is given by

F ′(x) = f ′(g(x)).g′(x)

Example: Findd

dx

(√x2 + 3x

)

Example: Findd

dx

(√x2 + 3x

)

The chain rule says if

F (x) = f(g(x)) then F ′(x) = f ′(g(x)).g′(x)

Here we have the outer function f(x) =√x, and

the inner function g(x) = x2 + 3x. Using our rulesof differentiation we have

f ′(x) =1

2√x

and g′(x) = 2x+ 3.

To calculaute f ′(g(x)), we use g(x) as the inputinto the derivative function f ′(x). Therefore

f ′(g(x)) =1

2√x2 + 3x

Example: Findd

dx

(√x2 + 3x

)The chain rule says if

F (x) = f(g(x)) then F ′(x) = f ′(g(x)).g′(x)

Here we have the outer function f(x) =√x, and

the inner function g(x) = x2 + 3x. Using our rulesof differentiation we have

f ′(x) =1

2√x

and g′(x) = 2x+ 3.

To calculaute f ′(g(x)), we use g(x) as the inputinto the derivative function f ′(x). Therefore

f ′(g(x)) =1

2√x2 + 3x

Example: Findd

dx

(√x2 + 3x

)The chain rule says if

F (x) = f(g(x)) then F ′(x) = f ′(g(x)).g′(x)

Here we have the outer function f(x) =√x, and

the inner function g(x) = x2 + 3x.

Using our rulesof differentiation we have

f ′(x) =1

2√x

and g′(x) = 2x+ 3.

To calculaute f ′(g(x)), we use g(x) as the inputinto the derivative function f ′(x). Therefore

f ′(g(x)) =1

2√x2 + 3x

Example: Findd

dx

(√x2 + 3x

)The chain rule says if

F (x) = f(g(x)) then F ′(x) = f ′(g(x)).g′(x)

Here we have the outer function f(x) =√x, and

the inner function g(x) = x2 + 3x. Using our rulesof differentiation we have

f ′(x) =1

2√x

and g′(x) = 2x+ 3.

To calculaute f ′(g(x)), we use g(x) as the inputinto the derivative function f ′(x). Therefore

f ′(g(x)) =1

2√x2 + 3x

Example: Findd

dx

(√x2 + 3x

)The chain rule says if

F (x) = f(g(x)) then F ′(x) = f ′(g(x)).g′(x)

Here we have the outer function f(x) =√x, and

the inner function g(x) = x2 + 3x. Using our rulesof differentiation we have

f ′(x) =1

2√x

and g′(x) = 2x+ 3.

To calculaute f ′(g(x)), we use g(x) as the inputinto the derivative function f ′(x). Therefore

f ′(g(x)) =1

2√x2 + 3x

If F (x) = f (g(x)) then

F ′(x) = f ′(g(x)).g′(x)

Example: Findd

dx

(√x2 + 3x

)

Solution:

d

dx

(√x2 + 3x

)=

1

2√x2 + 3x

(2x + 3)

=2x + 3

2√x2 + 3x

If F (x) = f (g(x)) then

F ′(x) = f ′(g(x)).g′(x)

Example: Findd

dx

(√x2 + 3x

)Solution:

d

dx

(√x2 + 3x

)=

1

2√x2 + 3x

(2x + 3)

=2x + 3

2√x2 + 3x

Chain rule examples:

Differentiate the following functions:

1. H(x) = sin(x2)

2. T (x) = sin2 x

The key to successfully applying the Chain

Rule is recognising the inner and outer

function.

The easiest way to do this is to think about

how you would evaluate the function at a

given x value.

Chain rule examples:

Differentiate the following functions:

1. H(x) = sin(x2)

2. T (x) = sin2 x

The key to successfully applying the Chain

Rule is recognising the inner and outer

function.

The easiest way to do this is to think about

how you would evaluate the function at a

given x value.

Chain rule examples:

Differentiate the following functions:

1. H(x) = sin(x2)

2. T (x) = sin2 x

The key to successfully applying the Chain

Rule is recognising the inner and outer

function.

The easiest way to do this is to think about

how you would evaluate the function at a

given x value.

1. H(x) = sin(x2)

2. T (x) = sin2 x = (sinx)2

How would you evaluate H(x) for some value

of x?

First you would square x and then you

would put the result into the sin function.

For T (x), you would first calculate sinx and

then square the result.

1. H(x) = sin(x2)

2. T (x) = sin2 x = (sinx)2

How would you evaluate H(x) for some value

of x? First you would square x and then you

would put the result into the sin function.

For T (x), you would first calculate sinx and

then square the result.

1. H(x) = sin(x2)

2. T (x) = sin2 x = (sinx)2

How would you evaluate H(x) for some value

of x? First you would square x and then you

would put the result into the sin function.

For T (x), you would first calculate sinx and

then square the result.

1. H(x) = sin(x2)

2. T (x) = sin2 x

Solutions:

1. We have H(x) = f (g(x)) where

g(x) = x2 and f (x) = sinx. Therefore

H ′(x) = f ′(g(x)).g′(x) = cos(x2).2x

2. Now we have T (x) = f (g(x)) where

g(x) = sinx and f (x) = x2. Therefore

T ′(x) = f ′(g(x)).g′(x)

= 2(sinx)1. cosx = 2 sinx cosx

1. H(x) = sin(x2)

2. T (x) = sin2 x

Solutions:

1. We have H(x) = f (g(x)) where

g(x) = x2 and f (x) = sinx.

Therefore

H ′(x) = f ′(g(x)).g′(x) = cos(x2).2x

2. Now we have T (x) = f (g(x)) where

g(x) = sinx and f (x) = x2. Therefore

T ′(x) = f ′(g(x)).g′(x)

= 2(sinx)1. cosx = 2 sinx cosx

1. H(x) = sin(x2)

2. T (x) = sin2 x

Solutions:

1. We have H(x) = f (g(x)) where

g(x) = x2 and f (x) = sinx. Therefore

H ′(x) = f ′(g(x)).g′(x) = cos(x2).2x

2. Now we have T (x) = f (g(x)) where

g(x) = sinx and f (x) = x2. Therefore

T ′(x) = f ′(g(x)).g′(x)

= 2(sinx)1. cosx = 2 sinx cosx

1. H(x) = sin(x2)

2. T (x) = sin2 x

Solutions:

1. We have H(x) = f (g(x)) where

g(x) = x2 and f (x) = sinx. Therefore

H ′(x) = f ′(g(x)).g′(x) = cos(x2).2x

2. Now we have T (x) = f (g(x)) where

g(x) = sinx and f (x) = x2.

Therefore

T ′(x) = f ′(g(x)).g′(x)

= 2(sinx)1. cosx = 2 sinx cosx

1. H(x) = sin(x2)

2. T (x) = sin2 x

Solutions:

1. We have H(x) = f (g(x)) where

g(x) = x2 and f (x) = sinx. Therefore

H ′(x) = f ′(g(x)).g′(x) = cos(x2).2x

2. Now we have T (x) = f (g(x)) where

g(x) = sinx and f (x) = x2. Therefore

T ′(x) = f ′(g(x)).g′(x)

= 2(sinx)1. cosx = 2 sinx cosx

1. H(x) = sin(x2)

2. T (x) = sin2 x

Solutions:

1. We have H(x) = f (g(x)) where

g(x) = x2 and f (x) = sinx. Therefore

H ′(x) = f ′(g(x)).g′(x) = cos(x2).2x

2. Now we have T (x) = f (g(x)) where

g(x) = sinx and f (x) = x2. Therefore

T ′(x) = f ′(g(x)).g′(x)

= 2(sinx)1. cosx = 2 sinx cosx

More chain rule examples:

Differentiate the following functions:

3. y = (x3 − 1)100

4. f (x) =1

3√x2 + x + 1

5. g(t) =

(t− 1

2t + 1

)9

Try them first on your own. The solutions

are on the slides that follow.

3. y = (x3 − 1)100

y′ = 100(x3−1)99 ddx

(x3−1)

= 100(x3 − 1)993x2 = 300x2(x3 − 1)99

4. f (x) =1

3√x2 + x + 1

= (x2 + x + 1)−1/3

f ′(x) = (−1/3)(x2+x+1)−4/3 ddx

(x2+x+1)

=−1

3(x2 + x + 1)4/3(2x + 1) =

−2x− 1

3(x2 + x + 1)4/3

3. y = (x3 − 1)100

y′ = 100(x3−1)99 ddx

(x3−1)

= 100(x3 − 1)993x2 = 300x2(x3 − 1)99

4. f (x) =1

3√x2 + x + 1

= (x2 + x + 1)−1/3

f ′(x) = (−1/3)(x2+x+1)−4/3 ddx

(x2+x+1)

=−1

3(x2 + x + 1)4/3(2x + 1) =

−2x− 1

3(x2 + x + 1)4/3

3. y = (x3 − 1)100

y′ = 100(x3−1)99 ddx

(x3−1)

= 100(x3 − 1)993x2 =

300x2(x3 − 1)99

4. f (x) =1

3√x2 + x + 1

= (x2 + x + 1)−1/3

f ′(x) = (−1/3)(x2+x+1)−4/3 ddx

(x2+x+1)

=−1

3(x2 + x + 1)4/3(2x + 1) =

−2x− 1

3(x2 + x + 1)4/3

3. y = (x3 − 1)100

y′ = 100(x3−1)99 ddx

(x3−1)

= 100(x3 − 1)993x2 = 300x2(x3 − 1)99

4. f (x) =1

3√x2 + x + 1

= (x2 + x + 1)−1/3

f ′(x) = (−1/3)(x2+x+1)−4/3 ddx

(x2+x+1)

=−1

3(x2 + x + 1)4/3(2x + 1) =

−2x− 1

3(x2 + x + 1)4/3

3. y = (x3 − 1)100

y′ = 100(x3−1)99 ddx

(x3−1)

= 100(x3 − 1)993x2 = 300x2(x3 − 1)99

4. f (x) =1

3√x2 + x + 1

= (x2 + x + 1)−1/3

f ′(x) = (−1/3)(x2+x+1)−4/3 ddx

(x2+x+1)

=−1

3(x2 + x + 1)4/3(2x + 1) =

−2x− 1

3(x2 + x + 1)4/3

3. y = (x3 − 1)100

y′ = 100(x3−1)99 ddx

(x3−1)

= 100(x3 − 1)993x2 = 300x2(x3 − 1)99

4. f (x) =1

3√x2 + x + 1

= (x2 + x + 1)−1/3

f ′(x) = (−1/3)(x2+x+1)−4/3 ddx

(x2+x+1)

=−1

3(x2 + x + 1)4/3(2x + 1) =

−2x− 1

3(x2 + x + 1)4/3

3. y = (x3 − 1)100

y′ = 100(x3−1)99 ddx

(x3−1)

= 100(x3 − 1)993x2 = 300x2(x3 − 1)99

4. f (x) =1

3√x2 + x + 1

= (x2 + x + 1)−1/3

f ′(x) = (−1/3)(x2+x+1)−4/3 ddx

(x2+x+1)

=−1

3(x2 + x + 1)4/3(2x + 1) =

−2x− 1

3(x2 + x + 1)4/3

3. y = (x3 − 1)100

y′ = 100(x3−1)99 ddx

(x3−1)

= 100(x3 − 1)993x2 = 300x2(x3 − 1)99

4. f (x) =1

3√x2 + x + 1

= (x2 + x + 1)−1/3

f ′(x) = (−1/3)(x2+x+1)−4/3 ddx

(x2+x+1)

=−1

3(x2 + x + 1)4/3(2x + 1)

=−2x− 1

3(x2 + x + 1)4/3

3. y = (x3 − 1)100

y′ = 100(x3−1)99 ddx

(x3−1)

= 100(x3 − 1)993x2 = 300x2(x3 − 1)99

4. f (x) =1

3√x2 + x + 1

= (x2 + x + 1)−1/3

f ′(x) = (−1/3)(x2+x+1)−4/3 ddx

(x2+x+1)

=−1

3(x2 + x + 1)4/3(2x + 1) =

−2x− 1

3(x2 + x + 1)4/3

5. g(t) =

(t− 1

2t + 1

)9

g′(t) = 9

(t− 1

2t + 1

)8d

dt

(t− 1

2t + 1

)

= 9

(t− 1

2t + 1

)8((1)(2t + 1)− (t− 1)(2)

(2t + 1)2

)

= 9

(t− 1

2t + 1

)8(3

(2t + 1)2

)=

27(t− 1)8

(2t + 1)10

5. g(t) =

(t− 1

2t + 1

)9

g′(t) = 9

(t− 1

2t + 1

)8d

dt

(t− 1

2t + 1

)

= 9

(t− 1

2t + 1

)8((1)(2t + 1)− (t− 1)(2)

(2t + 1)2

)

= 9

(t− 1

2t + 1

)8(3

(2t + 1)2

)=

27(t− 1)8

(2t + 1)10

5. g(t) =

(t− 1

2t + 1

)9

g′(t) = 9

(t− 1

2t + 1

)8d

dt

(t− 1

2t + 1

)

= 9

(t− 1

2t + 1

)8((1)(2t + 1)− (t− 1)(2)

(2t + 1)2

)

= 9

(t− 1

2t + 1

)8(3

(2t + 1)2

)=

27(t− 1)8

(2t + 1)10

5. g(t) =

(t− 1

2t + 1

)9

g′(t) = 9

(t− 1

2t + 1

)8d

dt

(t− 1

2t + 1

)

= 9

(t− 1

2t + 1

)8((1)(2t + 1)− (t− 1)(2)

(2t + 1)2

)

= 9

(t− 1

2t + 1

)8(3

(2t + 1)2

)

=27(t− 1)8

(2t + 1)10

5. g(t) =

(t− 1

2t + 1

)9

g′(t) = 9

(t− 1

2t + 1

)8d

dt

(t− 1

2t + 1

)

= 9

(t− 1

2t + 1

)8((1)(2t + 1)− (t− 1)(2)

(2t + 1)2

)

= 9

(t− 1

2t + 1

)8(3

(2t + 1)2

)=

27(t− 1)8

(2t + 1)10

Another method for trig derivatives

Findd

dx(cscx)

using the fact that cscx = (sinx)−1.

Try this on your own and then look at the

solution on the next slide.

d

dx(cscx) =

d

dx

((sinx)−1

)

The outer function is f (x) = x−1 and the

inner function is g(x) = sinx.

d

dxf (g(x)) = f ′(g(x)).g′(x)

Therefored

dx

((sinx)−1

)= (−1)(sinx)−2 d

dx(sinx)

=−1

(sinx)2(cosx) =

− cosx

sinx· 1

sinx= −cotx cscx

d

dx(cscx) =

d

dx

((sinx)−1

)The outer function is f (x) = x−1 and the

inner function is g(x) = sinx.

d

dxf (g(x)) = f ′(g(x)).g′(x)

Therefored

dx

((sinx)−1

)= (−1)(sinx)−2 d

dx(sinx)

=−1

(sinx)2(cosx) =

− cosx

sinx· 1

sinx= −cotx cscx

d

dx(cscx) =

d

dx

((sinx)−1

)The outer function is f (x) = x−1 and the

inner function is g(x) = sinx.

d

dxf (g(x)) = f ′(g(x)).g′(x)

Therefored

dx

((sinx)−1

)= (−1)(sinx)−2 d

dx(sinx)

=−1

(sinx)2(cosx) =

− cosx

sinx· 1

sinx= −cotx cscx

d

dx(cscx) =

d

dx

((sinx)−1

)The outer function is f (x) = x−1 and the

inner function is g(x) = sinx.

d

dxf (g(x)) = f ′(g(x)).g′(x)

Therefored

dx

((sinx)−1

)= (−1)(sinx)−2 d

dx(sinx)

=−1

(sinx)2(cosx) =

− cosx

sinx· 1

sinx= −cotx cscx

d

dx(cscx) =

d

dx

((sinx)−1

)The outer function is f (x) = x−1 and the

inner function is g(x) = sinx.

d

dxf (g(x)) = f ′(g(x)).g′(x)

Therefored

dx

((sinx)−1

)= (−1)(sinx)−2 d

dx(sinx)

=−1

(sinx)2(cosx)

=− cosx

sinx· 1

sinx= −cotx cscx

d

dx(cscx) =

d

dx

((sinx)−1

)The outer function is f (x) = x−1 and the

inner function is g(x) = sinx.

d

dxf (g(x)) = f ′(g(x)).g′(x)

Therefored

dx

((sinx)−1

)= (−1)(sinx)−2 d

dx(sinx)

=−1

(sinx)2(cosx) =

− cosx

sinx· 1

sinx=

−cotx cscx

d

dx(cscx) =

d

dx

((sinx)−1

)The outer function is f (x) = x−1 and the

inner function is g(x) = sinx.

d

dxf (g(x)) = f ′(g(x)).g′(x)

Therefored

dx

((sinx)−1

)= (−1)(sinx)−2 d

dx(sinx)

=−1

(sinx)2(cosx) =

− cosx

sinx· 1

sinx= −cotx cscx

This is a good point to take a break. Have a

look over the examples that have been done

already and make sure you understand them.

Come back in 20 minutes, or even tomorrow,

to carry on.

A useful piece of terminology

If F (x) = f(g(x)) then F ′(x) = f ′(g(x)).g′(x)

The derivative of the inner function is often

referred to as the “chain rule factor”, e.g.:

d

dt(cot 2t) = (− csc2 2t)2 = −2 csc2 2t

Here the 2 is what we would call the chain

rule factor. When doing a long differentiation

problem, make sure that you remember to

include the chain rule factor in each step!

Chain rule with product rule

Differentiate

y = (2x + 1)5(x3 − x + 1)4

Try this on your own before looking at the

solution. Hint: apply the product rule first

(recall (fg)′ = f ′g + g′f) and then use the

chain rule to calculate f ′ and g′.

Tidy up your answer at the end by taking out

common factors.

Chain rule with product rule

Differentiate

y = (2x + 1)5(x3 − x + 1)4

Try this on your own before looking at the

solution. Hint: apply the product rule first

(recall (fg)′ = f ′g + g′f) and then use the

chain rule to calculate f ′ and g′.

Tidy up your answer at the end by taking out

common factors.

y′ =d

dx

((2x + 1)5

)(x3 − x + 1)4+

(2x + 1)5d

dx

((x3 − x + 1)4

)

=

(5(2x + 1)4 · d

dx(2x + 1)

)(x3−x+1)4+

(2x+1)5(

(4(x3 − x + 1)3 · d

dx(x3 − x + 1)

)=(5(2x + 1)4 · 2

)(x3 − x + 1)4+

(2x + 1)5(4(x3 − x + 1)3 · (3x2 − 1)

)

y′ =d

dx

((2x + 1)5

)(x3 − x + 1)4+

(2x + 1)5d

dx

((x3 − x + 1)4

)=

(5(2x + 1)4 · d

dx(2x + 1)

)(x3−x+1)4+

(2x+1)5(

(4(x3 − x + 1)3 · d

dx(x3 − x + 1)

)

=(5(2x + 1)4 · 2

)(x3 − x + 1)4+

(2x + 1)5(4(x3 − x + 1)3 · (3x2 − 1)

)

y′ =d

dx

((2x + 1)5

)(x3 − x + 1)4+

(2x + 1)5d

dx

((x3 − x + 1)4

)=

(5(2x + 1)4 · d

dx(2x + 1)

)(x3−x+1)4+

(2x+1)5(

(4(x3 − x + 1)3 · d

dx(x3 − x + 1)

)=(5(2x + 1)4 · 2

)(x3 − x + 1)4+

(2x + 1)5(4(x3 − x + 1)3 · (3x2 − 1)

)

= 10(2x + 1)4(x3 − x + 1)4+

4(2x + 1)5(x3 − x + 1)3(3x2 − 1)

= 2(2x+1)4(x3− x+1)3(5(x3− x+1)+

2(2x + 1)(3x2 − 1))

= 2(2x+1)4(x3−x+1)3(17x3+6x2−9x+3)

= 10(2x + 1)4(x3 − x + 1)4+

4(2x + 1)5(x3 − x + 1)3(3x2 − 1)

= 2(2x+1)4(x3− x+1)3(5(x3− x+1)+

2(2x + 1)(3x2 − 1))

= 2(2x+1)4(x3−x+1)3(17x3+6x2−9x+3)

= 10(2x + 1)4(x3 − x + 1)4+

4(2x + 1)5(x3 − x + 1)3(3x2 − 1)

= 2(2x+1)4(x3− x+1)3(5(x3− x+1)+

2(2x + 1)(3x2 − 1))

= 2(2x+1)4(x3−x+1)3(17x3+6x2−9x+3)

Combining other rules

We can use the chain rule with the rule for

the derivative of ex.

Find the derivative:

h(x) = esinx

We can think of h(x) = f (g(x)) where

g(x) = sinx and f (x) = ex. Recall thatddx(e

x) = ex. Therefore

h′(x) = f ′(g(x)) · g′(x) = esinx cosx.

Combining other rules

We can use the chain rule with the rule for

the derivative of ex.

Find the derivative:

h(x) = esinx

We can think of h(x) = f (g(x)) where

g(x) = sinx and f (x) = ex. Recall thatddx(e

x) = ex. Therefore

h′(x) = f ′(g(x)) · g′(x) = esinx cosx.

Combining other rules

We can use the chain rule with the rule for

the derivative of ex.

Find the derivative:

h(x) = esinx

We can think of h(x) = f (g(x)) where

g(x) = sinx and f (x) = ex.

Recall thatddx(e

x) = ex. Therefore

h′(x) = f ′(g(x)) · g′(x) = esinx cosx.

Combining other rules

We can use the chain rule with the rule for

the derivative of ex.

Find the derivative:

h(x) = esinx

We can think of h(x) = f (g(x)) where

g(x) = sinx and f (x) = ex. Recall thatddx(e

x) = ex.

Therefore

h′(x) = f ′(g(x)) · g′(x) = esinx cosx.

Combining other rules

We can use the chain rule with the rule for

the derivative of ex.

Find the derivative:

h(x) = esinx

We can think of h(x) = f (g(x)) where

g(x) = sinx and f (x) = ex. Recall thatddx(e

x) = ex. Therefore

h′(x) = f ′(g(x)) · g′(x) =

esinx cosx.

Combining other rules

We can use the chain rule with the rule for

the derivative of ex.

Find the derivative:

h(x) = esinx

We can think of h(x) = f (g(x)) where

g(x) = sinx and f (x) = ex. Recall thatddx(e

x) = ex. Therefore

h′(x) = f ′(g(x)) · g′(x) = esinx cosx.

A reminder about ex and lnx

Two important facts:

ln e = 1 and elnx = x.

How do we get these?

We have loga a = 1 for any a > 0 and therefore

ln e = loge e = 1.

Also, for a > 0 and b > 0 we have

loga b = c ⇐⇒ ac = b.

Now lnx = lnx so loge x = loge x. Henceeloge x = x. This gives us elnx = x.

A reminder about ex and lnx

Two important facts:

ln e = 1 and elnx = x.

How do we get these?

We have loga a = 1 for any a > 0 and therefore

ln e = loge e = 1.

Also, for a > 0 and b > 0 we have

loga b = c ⇐⇒ ac = b.

Now lnx = lnx so loge x = loge x. Henceeloge x = x. This gives us elnx = x.

A reminder about ex and lnx

Two important facts:

ln e = 1 and elnx = x.

How do we get these?

We have loga a = 1 for any a > 0 and therefore

ln e = loge e = 1.

Also, for a > 0 and b > 0 we have

loga b = c ⇐⇒ ac = b.

Now lnx = lnx so loge x = loge x. Henceeloge x = x. This gives us elnx = x.

A reminder about ex and lnx

Two important facts:

ln e = 1 and elnx = x.

How do we get these?

We have loga a = 1 for any a > 0 and therefore

ln e = loge e = 1.

Also, for a > 0 and b > 0 we have

loga b = c ⇐⇒ ac = b.

Now lnx = lnx so loge x = loge x. Henceeloge x = x. This gives us elnx = x.

Derivatives of exponential functions:

In Ch 3.1 we calculated the following:

d

dxax = ax lim

h→0

ah − 1

h= ax.f ′(0)

Now we can show that for a > 0 we have

d

dxax = ax ln a

We will use the fact that a = eln a and so

ax =(eln a)x

= e(ln a)x

Derivatives of exponential functions:

In Ch 3.1 we calculated the following:

d

dxax = ax lim

h→0

ah − 1

h= ax.f ′(0)

Now we can show that for a > 0 we have

d

dxax = ax ln a

We will use the fact that a = eln a and so

ax =(eln a)x

= e(ln a)x

Derivatives of exponential functions:

In Ch 3.1 we calculated the following:

d

dxax = ax lim

h→0

ah − 1

h= ax.f ′(0)

Now we can show that for a > 0 we have

d

dxax = ax ln a

We will use the fact that a = eln a and so

ax =(eln a)x

= e(ln a)x

d

dx(ax) =

d

dx

(e(ln a)x

)

= e(ln a)x · ddx

((ln a)x)

=(eln a)x · (ln a) d

dx(x)

=(eln a)x · (ln a) · (1)

= ax ln a

d

dx(ax) =

d

dx

(e(ln a)x

)= e(ln a)x · d

dx((ln a)x)

=(eln a)x · (ln a) d

dx(x)

=(eln a)x · (ln a) · (1)

= ax ln a

d

dx(ax) =

d

dx

(e(ln a)x

)= e(ln a)x · d

dx((ln a)x)

=(eln a)x · (ln a) d

dx(x)

=(eln a)x · (ln a) · (1)

= ax ln a

d

dx(ax) =

d

dx

(e(ln a)x

)= e(ln a)x · d

dx((ln a)x)

=(eln a)x · (ln a) d

dx(x)

=(eln a)x · (ln a) · (1)

= ax ln a

d

dx(ax) =

d

dx

(e(ln a)x

)= e(ln a)x · d

dx((ln a)x)

=(eln a)x · (ln a) d

dx(x)

=(eln a)x · (ln a) · (1)

= ax ln a

Exponential function examples

Differentiate

y = 2x

We get y′ = 2x ln 2.

Now for something harder: find F ′(x) if

F (x) = 35x

See next slide for solution(s).

Exponential function examples

Differentiate

y = 2x

We get y′ = 2x ln 2.

Now for something harder: find F ′(x) if

F (x) = 35x

See next slide for solution(s).

Exponential function examples

Differentiate

y = 2x

We get y′ = 2x ln 2.

Now for something harder: find F ′(x) if

F (x) = 35x

See next slide for solution(s).

Find F ′(x) if F (x) = 35x.

Solution 1:

Note that F (x) = f (g(x)) where

g(x) = 5x and f (x) = 3x. Now g′(x) = 5

and f ′(x) = 3x ln 3. Hence

F ′(x) = f ′(g(x)) · g′(x) = (35x ln 3)5.

Solution 2 (from a student last year):

F (x) = (35)x so, using the fact thatddx(a

x) = ax ln a, we get

F ′(x) = (35)x ln(35) = (35x) ln(35) = (35x)5 ln 3.

Find F ′(x) if F (x) = 35x.

Solution 1: Note that F (x) = f (g(x)) where

g(x) = 5x and f (x) = 3x.

Now g′(x) = 5

and f ′(x) = 3x ln 3. Hence

F ′(x) = f ′(g(x)) · g′(x) = (35x ln 3)5.

Solution 2 (from a student last year):

F (x) = (35)x so, using the fact thatddx(a

x) = ax ln a, we get

F ′(x) = (35)x ln(35) = (35x) ln(35) = (35x)5 ln 3.

Find F ′(x) if F (x) = 35x.

Solution 1: Note that F (x) = f (g(x)) where

g(x) = 5x and f (x) = 3x. Now g′(x) = 5

and f ′(x) = 3x ln 3.

Hence

F ′(x) = f ′(g(x)) · g′(x) = (35x ln 3)5.

Solution 2 (from a student last year):

F (x) = (35)x so, using the fact thatddx(a

x) = ax ln a, we get

F ′(x) = (35)x ln(35) = (35x) ln(35) = (35x)5 ln 3.

Find F ′(x) if F (x) = 35x.

Solution 1: Note that F (x) = f (g(x)) where

g(x) = 5x and f (x) = 3x. Now g′(x) = 5

and f ′(x) = 3x ln 3. Hence

F ′(x) = f ′(g(x)) · g′(x) = (35x ln 3)5.

Solution 2 (from a student last year):

F (x) = (35)x so, using the fact thatddx(a

x) = ax ln a, we get

F ′(x) = (35)x ln(35) = (35x) ln(35) = (35x)5 ln 3.

Find F ′(x) if F (x) = 35x.

Solution 1: Note that F (x) = f (g(x)) where

g(x) = 5x and f (x) = 3x. Now g′(x) = 5

and f ′(x) = 3x ln 3. Hence

F ′(x) = f ′(g(x)) · g′(x) = (35x ln 3)5.

Solution 2 (from a student last year):

F (x) = (35)x so, using the fact thatddx(a

x) = ax ln a, we get

F ′(x) = (35)x ln(35) = (35x) ln(35) = (35x)5 ln 3.

Find F ′(x) if F (x) = 35x.

Solution 1: Note that F (x) = f (g(x)) where

g(x) = 5x and f (x) = 3x. Now g′(x) = 5

and f ′(x) = 3x ln 3. Hence

F ′(x) = f ′(g(x)) · g′(x) = (35x ln 3)5.

Solution 2 (from a student last year):

F (x) = (35)x so, using the fact thatddx(a

x) = ax ln a, we get

F ′(x) = (35)x ln(35) = (35x) ln(35) = (35x)5 ln 3.

Double chain rule! We can apply the

chain rule to composite functions that are

made by composing more than two functions.

Examples:

1. j(x) = sin(cos(tanx))

2. y = esec 3θ

Try these on your own and then look at the

solutions. Hint: if K(x) = f (g(h(x))) then

K ′(x) = f ′(g(h(x))) · ddx

(g(h(x)))

= f ′(g(h(x))) · g′(h(x)) · h′(x)

Double chain rule! We can apply the

chain rule to composite functions that are

made by composing more than two functions.

Examples:

1. j(x) = sin(cos(tanx))

2. y = esec 3θ

Try these on your own and then look at the

solutions.

Hint: if K(x) = f (g(h(x))) then

K ′(x) = f ′(g(h(x))) · ddx

(g(h(x)))

= f ′(g(h(x))) · g′(h(x)) · h′(x)

Double chain rule! We can apply the

chain rule to composite functions that are

made by composing more than two functions.

Examples:

1. j(x) = sin(cos(tanx))

2. y = esec 3θ

Try these on your own and then look at the

solutions. Hint: if K(x) = f (g(h(x))) then

K ′(x) = f ′(g(h(x))) · ddx

(g(h(x)))

= f ′(g(h(x))) · g′(h(x)) · h′(x)

Solutions:

1. j′(x) = cos(cos(tanx))d

dx

(cos(tanx)

)= cos(cos(tanx))

(− sin(tanx)

) ddx

(tanx)

= − cos(cos(tanx)) sin(tanx) sec2 x

2. y′ = esec 3θd

dx

(sec 3θ

)= esec 3θ sec 3θ tan 3θ

d

dx(3θ)

= 3esec 3θ sec 3θ tan 3θ

Solutions:

1. j′(x) = cos(cos(tanx))d

dx

(cos(tanx)

)

= cos(cos(tanx))(− sin(tanx)

) ddx

(tanx)

= − cos(cos(tanx)) sin(tanx) sec2 x

2. y′ = esec 3θd

dx

(sec 3θ

)= esec 3θ sec 3θ tan 3θ

d

dx(3θ)

= 3esec 3θ sec 3θ tan 3θ

Solutions:

1. j′(x) = cos(cos(tanx))d

dx

(cos(tanx)

)= cos(cos(tanx))

(− sin(tanx)

) ddx

(tanx)

= − cos(cos(tanx)) sin(tanx) sec2 x

2. y′ = esec 3θd

dx

(sec 3θ

)= esec 3θ sec 3θ tan 3θ

d

dx(3θ)

= 3esec 3θ sec 3θ tan 3θ

Solutions:

1. j′(x) = cos(cos(tanx))d

dx

(cos(tanx)

)= cos(cos(tanx))

(− sin(tanx)

) ddx

(tanx)

= − cos(cos(tanx)) sin(tanx) sec2 x

2. y′ = esec 3θd

dx

(sec 3θ

)= esec 3θ sec 3θ tan 3θ

d

dx(3θ)

= 3esec 3θ sec 3θ tan 3θ

Solutions:

1. j′(x) = cos(cos(tanx))d

dx

(cos(tanx)

)= cos(cos(tanx))

(− sin(tanx)

) ddx

(tanx)

= − cos(cos(tanx)) sin(tanx) sec2 x

2. y′ = esec 3θd

dx

(sec 3θ

)

= esec 3θ sec 3θ tan 3θd

dx(3θ)

= 3esec 3θ sec 3θ tan 3θ

Solutions:

1. j′(x) = cos(cos(tanx))d

dx

(cos(tanx)

)= cos(cos(tanx))

(− sin(tanx)

) ddx

(tanx)

= − cos(cos(tanx)) sin(tanx) sec2 x

2. y′ = esec 3θd

dx

(sec 3θ

)= esec 3θ sec 3θ tan 3θ

d

dx(3θ)

= 3esec 3θ sec 3θ tan 3θ

Prescribed tut problems:

Complete the following exercises from the

8th edition (Ch 3.4):

1, 3, 5, 7, 9, 10, 11, 13, 17, 21, 28, 31, 47,

49, 51, 53, 61, 75, 80, 82, 84 (a & b)

A good understanding of the Chain Rule is

extremely important for understanding

the rest of the topics that we will cover this

year. That is why there are lots of prescribed

problems from this section.