Section 2.4 The Chain Rule

Section 2.4The Chain Rule

Unit 2 – Differentiation

Objectives:

1. Find the derivative of a composite function using the Chain Rule.2. Find the derivative of a function using the General Power Rule.3. Simplify the derivative of a function using algebra.4. Find the derivative of a trigonometric function using the Chain Rule.

The Chain Rule

The Chain Rule is an extremely important rule in calculus and allows us to extend differentiation to many more types of functions. Below are examples of functions that can be done without the Chain Rule, and those that can be done with it.

Without the Chain Rule With the Chain Rule

The Chain Rule expands differentiation to composite functions like or .

Theorem 2.10The Chain Rule

If is a differentiable function of and is a differentiable function of , then is a differentiable function of and

or, equivalently,

Decomposing Functions

» Part of the process in using the Chain Rule is figuring out the best way to decompose a function. Remember that a composite function is made from one function inside of another. For example, can be thought of as the function inside of the function , where

» Example 1 will illustrate this process further.

Example 1

» Decompose the following functions:𝑦= 𝑓 (𝑔 (𝑥 )) 𝑢=g(𝑥) 𝑦= 𝑓 (𝑢)

𝑢=2𝑥−3 𝑦=1𝑢

𝑢=3 𝑥

𝑢=5 𝑥2+3 𝑥

𝑢=cot (𝑥)

𝑦=cos (𝑢)

𝑦=√𝑢

𝑦=𝑢2

Example 2A

» Differentiate the following functions using methods learned in sections 2.2 & 2.3, and then by using the Chain Rule.

We will begin by multiplying out the entire polynomial:

𝑦 ′=𝐷𝑥 [𝑥3+6 𝑥2+12𝑥+8 ]=3 𝑥2+12𝑥+12Once multiplied out the derivative is easily completed, but getting there can be time consuming, especially for higher powers.

Example 2A continued…


We begin by decomposing the two functions like in Example 1:

Now for the Chain Rule:

From the Chain Rule, the derivative is:

So, to continue we must differentiate both of the decomposed functions.

Example 2A continued…


𝑢=𝑥+2By differentiating each we get:

So, the combined result is:

Which when multiplied out is (which isn’t always necessary):

𝑦 ′=3 (𝑥+2 )2=3 (𝑥2+4 𝑥+4 )=3 𝑥2+12 𝑥+12

𝑓 (𝑢)=𝑢3𝒖=𝒙+𝟐

Example 2B


To differentiate this without the Chain Rule requires using a double-angle trigonometric identity with the Product Rule:

𝑦=sin (2 𝑥)=2sin (𝑥 ) cos (𝑥)

Example 2B continued…


By decomposition we get:


By differentiating we get:

So, the result is:

Which by another double-angle identity is equivalent to:

𝑦 ′=2cos (2 𝑥 )=2 [cos2 (𝑥 )− sin2(𝑥 )]

Example 2C


We first differentiate using the Quotient Rule:

Example 2C continued…


𝑔 (𝑥 )=𝑢=3 𝑥+1

By decomposition we get:

𝑓 (𝑢)= 2𝑢


By differentiating we get:

𝑓 ′ (𝑢 )=2𝑢− 1=−2𝑢− 2=− 2𝑢2

𝑔 ′ (𝑥)=𝑢′=3

𝑦 ′= 𝑓 ′ (𝑢 )𝑢′=(− 2𝑢2 ) (3 )=− 6(3𝑥+1 )2

So, the result is:

Example 2 Summary

» In each case, we saw how three different problems can be solved with and without the Chain Rule. Once you get used to the Chain Rule, it becomes much faster and easier, and often can be done in your head. It also is convenient to serve as a way to check answers when a different method is requested.

Theorem 2.11The General Power Rule

If where is a differentiable function of and is a rational number, then

or, equivalently,

Example 3A

A. Find for

𝑑𝑦𝑑𝑥=3 (2 𝑥2+5 )2 (4 𝑥 )=12𝑥 (2𝑥2+5 )2¿𝑑𝑦

𝑑𝑢{ ¿

𝑑𝑢𝑑𝑥

𝑢′= 𝑑𝑢𝑑𝑥=4 𝑥

𝑦 ′=𝑑𝑦𝑑𝑢=3𝑢2

𝑑𝑦𝑑𝑥 =

𝑑𝑦𝑑𝑢 ∙

𝑑𝑢𝑑𝑥

Example 3B

B. If and then the derivative of is

This question is similar to a possible AP multiple choice question. It is unique in that it separates the functions being composed.

Example 3B

B. If and then find the derivative of

We will start by saying:

However, we will not use the Chain Rule on and , but rather make a different decomposition.



𝑓 (𝑔 (𝑥 ) )=h (𝑥 )= 1𝑥+1

𝑢=𝑥+1

𝑢′= 𝑑𝑢𝑑𝑥=1

h (𝑢)=1𝑢



𝑓 (𝑔 (𝑥 ) )=h (𝑥 )= 1𝑥+1 𝑢=𝑥+1

𝑑𝑢𝑑𝑥=1

𝑑 h𝑑𝑢=− 1

𝑢2

h (𝑢)=1𝑢

𝑑 h𝑑𝑥=

h𝑑𝑑𝑢 ∙

𝑑𝑢𝑑𝑥=− 1

𝑢2(1 )=− 1

𝑢2=− 1

(𝑥+1 )2

We are now ready to answer the multiple choice question.

Example 3B

B. If and then the derivative of is 𝑑

𝑑𝑥 [ 𝑓 (𝑔 (𝑥 ))]=− 1(𝑥+1 )2

Even if you solved the problem correctly, you could easily miss this question by not paying attention. Notice that d) looks like the correct answer, but it lacks the negative sign. So, the correct answer is not in the same format, but by looking closely we see the answer is b).

Example 3C

C. If and then equals

𝑑𝑑𝑥 𝑓 (𝑥 )= 𝑓 ′(𝑥 )=𝑔 (𝑥)

Extending the Chain Rule to More Functions

» Using Leibniz’s Notation in the previous example shows us how the Chain Rule can be extended to more functions.

» The next example will demonstrate this further.

𝑑𝑦𝑑𝑥 =

𝑑𝑦𝑑𝑢 ∙

𝑑𝑢𝑑𝑣 ∙

𝑑𝑣𝑑𝑥

Example 4

» Differentiate using the Chain Rule twice.

𝑦=(1+(𝑥2−5 )15 )10

𝑑𝑣𝑑𝑥=2𝑥 𝑑𝑢𝑑𝑣=15𝑣14 𝑑𝑦𝑑𝑢=10𝑢9

Example 5

» Find all points on the graph of for which and those for which does not exist.

𝑓 (𝑥)=3√ (𝑥2−4 )2 ¿ (𝑥2−4 )

23

𝑓 ′ (𝑥 ) ¿ 23

(𝑥2−4 )−13 (2 𝑥 )= 4 𝑥

3 3√𝑥2−4Since the derivative is a rational function, to find where , we set the numerator equal to zero and solve. To find where does not exist, we set the denominator equal to zero and solve.

Example 5 continued…

» Find all points on the graph of for which and those for which does not exist.

𝑓 ′ (𝑥 )= 4 𝑥3 3√𝑥2−4

:

:


» By studying the graphs of the original function and its derivative, it becomes more apparent.

𝑓 ′ (𝑥 )= 4 𝑥3√𝑥2−4

𝑓 ′(0)=0𝑥≠±2

1 2 3 4 5 6 7 8 9 10–1–2–3–4–5–6–7–8–9–10 x

1

2

3

4

5

6

7

8

9

10

–1

–2

–3

–4

–5

–6

–7

–8

–9

–10

y

𝑓 ′(𝑥 )

𝑓 (𝑥)

𝑓 (𝑥)=3√ (𝑥2−4 )2

You may recall that function is NOT differentiable at a sharp corner. This is indicative of the vertical asymptotes at .

Also note the slope of the blue graph is zero precisely where the green crosses the x-axis, at .

Example 6A

A. Differentiate .

Here we see it is much easier to apply the Chain Rule and differentiate than to use the Quotient Rule. On the next slide we will modify the function slightly, and still use the Chain Rule.

Example 6B

B. Differentiate .

𝑔 (𝑥 )=3 𝑥 (5−3 𝑥 )− 2Because the numerator is not a constant, if we try to apply the Chain Rule here, we will also have to use the Product Rule.


B. Differentiate .

Now to differentiate it with the Quotient Rule.

Clearly for this particular problem, the Chain Rule combined with the Product Rule was easier. Sometimes if you get stuck, try a different method.

Simplifying Derivatives

» On the last example we just saw a problem whose derivative was given as two different fractions:

» Many times though, for instance on the multiple choice section of the AP test, an answer may be given as one fraction only. The following three examples illustrate techniques for simplifying derivatives.

» Note, an alternative method would be to get a common denominator after getting two separate fractions.

𝑔 ′ (𝑥 )= 3(5−3 𝑥 )2

+18 𝑥

(5−3 𝑥 )3

Example 7

» Differentiate and simplify by factoring out the least powers.

𝑓 (𝑥)=2 𝑥2√9−𝑥2¿2 𝑥2 (9−𝑥2 )12

𝑓 𝑔

𝑓 ′ (𝑥 )=4 𝑥 (9−𝑥2 )12+2𝑥2( 12 ) (9− 𝑥2 )

− 12 (−2𝑥 )

𝑓 ′ 𝑔 𝑓 𝑔 ′¿¿{¿ {¿𝑢𝑢 ′Observe, when multiplying like bases,

add the exponents. We will use this result backwards to factor.

(9− 𝑥2 )− 12 (9−𝑥2)1=(9−𝑥2 )

− 12 (9−𝑥2 )22=(9−𝑥2)

12

+ =


» We will now factor out the least power, as well as the GCF.

𝑓 ′ (𝑥 )=4 𝑥 (9−𝑥2 )12+2𝑥2( 12 ) (9− 𝑥2 )

−𝟏𝟐 (−2𝑥 )

(9− 𝑥2 )− 12 (𝟗− 𝒙𝟐 )𝟏=(9−𝑥2 )

12

¿2 𝑥 (9− 𝑥2 )− 𝟏𝟐 [2 (𝟗− 𝒙𝟐 )+𝑥 ( 12 ) (−2𝑥 )]{ ¿GCF


» Now to simplify.

Example 8

» Differentiate and simplify.

h (𝑥 )= 4 𝑥3√𝑥2+1

¿4 𝑥

(𝑥2+1 )13

𝑓 (𝑥)

𝑔 (𝑥)

¿𝑔2

h ′ (𝑥)=4 (𝑥2+1 )

13−4 𝑥 ( 13 ) (𝑥2+1 )

− 23 (2 𝑥 )

(𝑥2+1 )23

{ ¿ {¿ ¿𝑓 ′ 𝑔 ′𝑓𝑔

𝑢𝑢 ′


» Now to simplify the derivative.

h ′ (𝑥 )

Factored from top.4÷ 43=3

(𝑥2+1 )− 23 (𝑥2+1 )1=(𝑥2+1 )

− 23 (𝑥2+1 )33=(𝑥2+1 )

13

First observe, when multiplying like bases, add the exponents.

+

We will use this result backwards to factor the derivative.

¿ 43

(𝑥2+1 )− 23 [ 3𝑥

2+3−2𝑥2

(𝑥2+1 )23 ]

Example 9

» Differentiate and simplify.

¿−2 (1+𝑥 ) (𝑥2+2𝑥−2 )

(2+𝑥2 )3

Example 10

» Differentiate the following trigonometric functions.

𝑦 ′=−sin (5 𝑥 ) 𝑑𝑑𝑥 [5𝑥 ]=−sin (5 𝑥) (5 )=−5sin (5𝑥 )

𝑦 ′=sec2 (4 𝑥−1 ) 𝑑𝑑𝑥 [4 𝑥−1 ]=sec2 (4 𝑥−1 ) (4 )=4 sec2 (4 𝑥−1 )

Example 11

» Parentheses change a lot with respect to trigonometric functions. Keep that in mind when differentiating.

A.

𝑦 ′=cos 2𝑥3 𝑑𝑑𝑥 [2𝑥3 ]=cos2 𝑥3 (6 𝑥2 )=6 𝑥2 cos2𝑥3

𝑦 ′=(sin 2 ) 𝑑𝑑𝑥 [𝑥3 ]= (sin 2 ) (3 𝑥2 )=(3sin 2 )𝑥2

Here acts as a constant since the variable is not inside the sine function.



C.

𝑦 ′=cos 8 𝑥3 𝑑𝑑𝑥 [ 8𝑥3 ]=cos 8𝑥3 (24 𝑥2 )=24 𝑥2cos 8 𝑥3

The last step used a double-angle identity.



𝑦 ′=12 (sin 𝑥 )− 12 𝑑𝑑𝑥 [sin𝑥 ]= 1

2 (sin 𝑥 )12

∙cos 𝑥=cos 𝑥2√sin 𝑥

Example 12

» Some trigonometric functions require repeated use of the Chain Rule. Differentiate the following:

𝑓 (𝑡 )=cos32 𝑡¿ (cos 2𝑡 )3

¿3 ( cos2 𝑡 )2 𝑑𝑑𝑥 [cos2 𝑡 ])

¿3 ( cos2 𝑡 )2 (−sin 2 𝑡 ) 𝑑𝑑𝑥 [2 𝑡 ]

¿3 ( cos2 𝑡 )2 (−sin 2 𝑡 ) (2 )

¿−6cos22 𝑡 sin 2 𝑡

Example 13

» Find an equation of the tangent line to the graph of at the point . Then determine all values of in the interval at which the graph of has a horizontal tangent.

𝑓 ′ (𝑥 )=−3sin 𝑥+3cos3 𝑥

First find the derivative and substitute in to find the slope at that point.



𝑚= 𝑓 ′ (𝜋2 )=−3Now use point-slope form to write the equation substituting in the point and the slope we just found.



Horizontal tangents occur where the slope is zero, so one method to find them is to graph the derivative and find the zeros of the function.

𝑓 ′ (𝑥 )=−3sin 𝑥+3cos3 𝑥=0

Set the window sizes as shown to maximize the graph. Note the x-min and x-max are & .



Horizontal tangents occur where the slope is zero, so one method to find them is to graph the derivative and find the zeros of the function.

𝑓 ′ (𝑥 )=−3sin 𝑥+3cos3 𝑥=0

𝑥=𝜋8 ,5𝜋8 , 3𝜋4 ,

9𝜋8 , 13𝜋8 , 7𝜋4

With a little playing around with the trace feature, it becomes obvious that the zeros are certain multiples of

Example 14A-G

» For Example 14A-G, the following table of functions and their derivatives is given. Questions like these may be on the AP test.

2 1 5 4

3 2 3 3

5 3 1 2

10 4 0 1

Example 14A

A. If , then

2 1 5 4

3 2 3 3

5 3 1 2

10 4 0 1

First we must differentiate to find the general function

𝐴 ′= 𝑓 ′+2𝑔 ′

So,

Example 14B

B. If , then

2 1 5 4

3 2 3 3

5 3 1 2

10 4 0 1


𝐵 ′= 𝑓 ′𝑔+ 𝑓 𝑔 ′

So,

Example 14C

C. If , then

2 1 5 4

3 2 3 3

5 3 1 2

10 4 0 1


So,

Example 14D

D. If , then

2 1 5 4

3 2 3 3

5 3 1 2

10 4 0 1


So,

Example 14E

E. If , then

2 1 5 4

3 2 3 3

5 3 1 2

10 4 0 1


𝐾 ′=𝑓 ′𝑔− 𝑓𝑔 ′

𝑔2

So,

Example 14F

F. If , then

2 1 5 4

3 2 3 3

5 3 1 2

10 4 0 1


𝑀 ′ (𝑥 )= 𝑓 ′ (𝑔 (𝑥 ) )𝑔 ′ (𝑥)

So,

Example 14G

G. If , then

2 1 5 4

3 2 3 3

5 3 1 2

10 4 0 1


𝑃 ′ (𝑥)= 𝑓 ′ (𝑥3 )3 𝑥2

So,

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Section 2.4 The Chain Rule