Chain Rule: Problems and Solutions

This screen contains only a summary of how to use the Chain Rule to calculate derivatives, and free problems for you to practice, each with a complete solution immediately available. If you need to get this down quickly, and don't have time to work through the more comprehensive material earlier in this Chapter, this screen should help. These are also great problems to practice for an exam.

Chain Rule Summary

You can always access our Handy Table of Derivatives and Differentiation Rules via the Key Formulas menu item at the top of every page.

You must use the Chain rule to find the derivative of any function that is comprised of one function inside of another function.

For instance, $(𝑥 2 + 1) 7$ is comprised of the inner function $𝑥 2 + 1$ inside the outer function $(\dots) 7 .$

As another example, $𝑒 s i n 𝑥$ is comprised of the inner function $s i n 𝑥$ inside the outer function $𝑒 \dots .$

As yet another example, $l n (𝑡 3 - 2 𝑡 2 + 5)$ is comprised of the inner function $𝑡 3 - 2 𝑡 2 + 5$ inside the outer function $l n (\dots) .$

Since each of these functions is comprised of one function inside of another function — known as a composite function — we must use the Chain rule to find its derivative, as shown in the problems below.

$Tip icon$

How can I tell what the inner and outer functions are?

Here's a foolproof method: Imagine calculating the value of the function for a particular value of $𝑥$ and identify the steps you would take, because you'll always automatically start with the inner function and work your way out to the outer function.

For example, imagine computing $(𝑥 2 + 1) 7$ for $𝑥 = 3 .$ Without thinking about it, you would first calculate $𝑥 2 + 1$ (which equals $32 + 1 = 10$ ), so that's the inner function, guaranteed. Then you would next calculate $107,$ and so $(\dots) 7$ is the outer function.

This imaginary computational process works every time to identify correctly what the inner and outer functions are.

Chain Rule

Consider a composite function whose outer function is $𝑓 (𝑥)$ and whose inner function is $𝑔 (𝑥) .$ The composite function is thus $𝑓 (𝑔 (𝑥)) .$ Its derivative is given by:

[𝑓 (𝑔 (𝑥))]' = 𝑓' (𝑔 (𝑥)) \cdot 𝑔' (𝑥) = [d e r i v a t i v e o f t h e o u t e r f u n c t i o n, e v a l u a t e d a t t h e i n n e r f u n c t i o n] \times [d e r i v a t i v e o f t h e i n n e r f u n c t i o n]

Alternatively, if we write $𝑦 = 𝑓 (𝑢)$ and $𝑢 = 𝑔 (𝑥),$ then

𝑑 𝑦 𝑑 𝑥 = 𝑑 𝑦 𝑑 𝑢 \cdot 𝑑 𝑢 𝑑 𝑥

Informally:

𝑑 𝑓 𝑑 𝑥 = [𝑑 𝑓 𝑑 (s t u f f), w i t h t h e s a m e s t u f f i n s i d e] \times 𝑑 𝑑 𝑥 (s t u f f)

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Even though few people admit it, almost everyone thinks along the lines of the informal approach in the blue boxes above. We'll illustrate in the problems below.

Chain Rule Example #1

Differentiate $𝑓 (𝑥) = (𝑥 2 + 1) 7$ .

Solutions.

We'll solve this using three different approaches — but we encourage you to become comfortable with the third approach as quickly as possible, because that's the one you'll use to compute derivatives quickly as the course progresses.

• Solution 1.

Let's use the first form of the Chain rule above:

[𝑓 (𝑔 (𝑥))]' = 𝑓' (𝑔 (𝑥)) \cdot 𝑔' (𝑥) = [d e r i v a t i v e o f t h e o u t e r f u n c t i o n, e v a l u a t e d a t t h e i n n e r f u n c t i o n] \times [d e r i v a t i v e o f t h e i n n e r f u n c t i o n]

We have the outer function $𝑓 (𝑢) = 𝑢 7$ and the inner function $𝑢 = 𝑔 (𝑥) = 𝑥 2 + 1 .$

Then $𝑓' (𝑢) = 7 𝑢 6,$ and $𝑔' (𝑥) = 2 𝑥 .$ Then

𝑓' (𝑥) = 7 𝑢 6 \cdot 2 𝑥 = 7 (𝑥 2 + 1) 6 \cdot 2 𝑥 ✓

We could of course simplify the result algebraically to $14 𝑥 (𝑥 2 + 1) 2,$ but we're leaving the result as written to emphasize the Chain rule term $2 𝑥$ at the end.

• Solution 2.

Let's use the second form of the Chain rule above:

𝑑 𝑦 𝑑 𝑥 = 𝑑 𝑦 𝑑 𝑢 \cdot 𝑑 𝑢 𝑑 𝑥

We have $𝑦 = 𝑢 7$ and $𝑢 = 𝑥 2 + 1 .$

Then $𝑑 𝑦 𝑑 𝑢 = 7 𝑢 6,$ and $𝑑 𝑢 𝑑 𝑥 = 2 𝑥 .$ Hence

𝑑 𝑦 𝑑 𝑥 = 7 𝑢 6 \cdot 2 𝑥 = 7 (𝑥 2 + 1) 6 \cdot 2 𝑥 ✓

• Solution 3.

Let's use the third form of the Chain rule above:

$Tip icon$

Instead, you'll think something like: "The function is a bunch of stuff to the 7th power. So the derivative is 7 times that same stuff to the 6th power, times the derivative of that stuff."

𝑑 𝑓 𝑑 𝑥 = [𝑑 𝑓 𝑑 (s t u f f), w i t h t h e s a m e s t u f f i n s i d e] \times 𝑑 𝑑 𝑥 (s t u f f)

𝑓 (𝑥) = (s t u f f) 7; s t u f f = 𝑥 2 + 1 T h e n 𝑓 (𝑥) = 𝑑 𝑓 𝑑 𝑥 = 7 (s t u f f) 6 \cdot (𝑑 𝑑 𝑥 (𝑥 2 + 1)) = 7 (𝑥 2 + 1) 6 \cdot (2 𝑥) ✓

Note: You'd never actually write out "stuff = ...." Instead just hold in your head what that "stuff" is, and proceed to write down the required derivatives.

There are lots more completely solved example problems below!

The problems below are grouped for practice purposes to use the Chain rule and: [Power rule] [Exponentials] [Trig Functions] [Product rule & Quotient rule] [Chain rule multiple times] [More problems and University exam problems]

Chain Rule & Power Rule

I f 𝑓 (𝑥) = (s t u f f) 𝑛, t h e n 𝑑 𝑓 𝑑 𝑥 = 𝑛 (t h a t s t u f f) 𝑛 - 1 \cdot 𝑑 𝑑 𝑥 (t h a t s t u f f)

You'll usually see this written as

𝑑 𝑑 𝑥 (𝑢 𝑛) = 𝑛 𝑢 𝑛 - 1 \cdot 𝑑 𝑢 𝑑 𝑥

The following six problems illustrate.

Chain Rule Practice Problem #1

Given

𝑓 (𝑥) = (3 𝑥 2 - 4 𝑥 + 5) 8,

𝑓' (𝑥) =

(A) 8 (3 𝑥 2 - 4 𝑥 + 5) 7 (B) 8 (3 𝑥 2 - 4 𝑥 + 5) 7 \cdot (6 𝑥 - 4) (C) 8 (6 𝑥 - 4) 7

(D) (3 𝑥 2 - 4 𝑥 + 5) 8 (E) n o n e o f t h e s e

ABCDE

Chain Rule Practice Problem #2

Given $𝑓 (𝑥) = t a n 3 𝑥,$ $𝑓' (𝑥) =$

Hint: Recall $t a n 3 𝑥 = [t a n 𝑥] 3 .$ Also recall that $𝑑 𝑑 𝑥 t a n 𝑥 = s e c 2 𝑥 .$

(A) 3 s e c 4 𝑥 (B) 3 t a n 2 𝑥 (C) t a n 3 𝑥 s e c 2 𝑥

(D) 3 t a n 2 𝑥 \cdot s e c 2 𝑥 (E) n o n e o f t h e s e

ABCDE

Chain Rule Practice Problem #3

Given

𝑓 (𝑥) = (c o s 𝑥 - s i n 𝑥) - 2, 𝑓' (𝑥) =

(A) - 2 (c o s 𝑥 - s i n 𝑥) - 1 \cdot (- s i n 𝑥 - c o s 𝑥) (B) - 2 (c o s 𝑥 - s i n 𝑥) - 3

(C) - 2 (c o s 𝑥 - s i n 𝑥) - 3 \cdot (- s i n 𝑥 - c o s 𝑥) (D) - 2 (- s i n 𝑥 - c o s 𝑥) - 3 \cdot (c o s 𝑥 - s i n 𝑥)

(E) n o n e o f t h e s e

ABCDE

Chain Rule Practice Problem #4

Given

𝑓 (𝑥) = (𝑥 5 + 𝑒 𝑥) 99, 𝑓' (𝑥) =

(A) 99 (𝑥 5 + 𝑒 𝑥) 98 \cdot (5 𝑥 4 + 𝑒 𝑥) (B) 99 (𝑥 5 + 𝑒 𝑥) 98 (C) 99 (𝑥 5 + 𝑒 𝑥) 98 \cdot (5 𝑥 4 + 𝑒 𝑥 - 1)

(D) 99 (5 𝑥 4 + 𝑒 𝑥) 98 (E) n o n e o f t h e s e

ABCDE

Chain Rule Practice Problem #5

Given

𝑓 (𝑥) = \sqrt 𝑥 2 + 1, 𝑓' (𝑥) =

(A) 12 1 \sqrt 𝑥 2 + 1 \cdot (2 𝑥) (B) 12 1 \sqrt 𝑥 2 + 1 (C) 12 1 \sqrt 2 𝑥

(D) 12 \sqrt 𝑥 2 + 1 (E) n o n e o f t h e s e

ABCDE

Chain Rule Practice Problem #6

Given

𝑓 (𝑥) = \sqrt s i n 𝑥, 𝑓' (𝑥) =

(A) 12 1 \sqrt s i n 𝑥 (B) 12 \sqrt s i n 𝑥 \cdot c o s 𝑥 (C) 12 1 \sqrt s i n 𝑥 \cdot c o s 𝑥

(D) - 12 1 \sqrt s i n 𝑥 \cdot c o s 𝑥 (E) n o n e o f t h e s e

ABCDE

Chain Rule & Exponentials

I f 𝑓 (𝑥) = 𝑒 (s t u f f), t h e n 𝑑 𝑓 𝑑 𝑥 = 𝑒 (t h a t s t u f f) \cdot 𝑑 𝑑 𝑥 (t h a t s t u f f)

You'll usually see this written as

𝑑 𝑑 𝑥 𝑒 𝑢 = 𝑒 𝑢 \cdot 𝑑 𝑢 𝑑 𝑥

The next three problems illustrate.

Chain Rule Practice Problem #7

Given $𝑓 (𝑥) = 𝑒 s i n 𝑥, 𝑓' (𝑥) =$

(A) 𝑒 c o s 𝑥 c o s 𝑥 (B) 𝑒 (s i n 𝑥 - 1) 𝑒 c o s 𝑥 (C) 𝑒 - s i n 𝑥 c o s 𝑥

(D) 𝑒 s i n 𝑥 c o s 𝑥 (E) n o n e o f t h e s e

ABCDE

Chain Rule Practice Problem #8

Given

𝑓 (𝑥) = 𝑒 𝑥 2, 𝑓' (𝑥) =

(A) 𝑒 𝑥 2 𝑒 2 𝑥 (B) 𝑒 𝑥 2 \cdot 2 𝑥 (C) 𝑒 𝑥 2 (D) 𝑒 2 𝑥 (E) n o n e o f t h e s e

ABCDE

Chain Rule Practice Problem #9

Given

𝑓 (𝑥) = 𝑒 (𝑥 7 - 4 𝑥 3 + 𝑥), 𝑓' (𝑥) =

(A) 𝑒 (𝑥 7 - 4 𝑥 3 + 𝑥) \cdot (7 𝑥 6 - 12 𝑥 2) (B) 𝑒 (𝑥 7 - 4 𝑥 3 + 𝑥) \cdot (7 𝑥 6 - 12 𝑥 2 + 1) (C) 𝑒 (𝑥 7 - 4 𝑥 3 + 𝑥)

(D) 𝑒 (7 𝑥 6 - 12 𝑥 2 + 1) (E) n o n e o f t h e s e

ABCDE

Chain Rule & Trig Functions

I f 𝑓 (𝑥) = s i n (s t u f f), t h e n 𝑑 𝑓 𝑑 𝑥 = c o s (t h a t s t u f f) \cdot 𝑑 𝑑 𝑥 (t h a t s t u f f)

You'll usually see this written as

𝑑 𝑑 𝑥 s i n 𝑢 = c o s 𝑢 \cdot 𝑑 𝑢 𝑑 𝑥

- - -

I f 𝑓 (𝑥) = c o s (s t u f f), t h e n 𝑑 𝑓 𝑑 𝑥 = - s i n (t h a t s t u f f) \cdot 𝑑 𝑑 𝑥 (t h a t s t u f f)

You'll usually see this written as

𝑑 𝑑 𝑥 c o s 𝑢 = - s i n 𝑢 \cdot 𝑑 𝑢 𝑑 𝑥

- - -

I f 𝑓 (𝑥) = t a n (s t u f f), t h e n 𝑑 𝑓 𝑑 𝑥 = s e c 2 (t h a t s t u f f) \cdot 𝑑 𝑑 𝑥 (t h a t s t u f f)

You'll usually see this written as

𝑑 𝑑 𝑥 t a n 𝑢 = s e c 2 𝑢 \cdot 𝑑 𝑢 𝑑 𝑥

The next two problems illustrate.

Chain Rule Practice Problem #10

Given

𝑓 (𝑥) = s i n (2 𝑥), 𝑓' (𝑥) =

(A) c o s (2 𝑥) \cdot (2) (B) s i n (2 𝑥) \cdot (2) (C) - c o s (2 𝑥) \cdot (2)

(D) c o s (2) (E) n o n e o f t h e s e

ABCDE

Chain Rule Practice Problem #11

Given $𝑓 (𝑥) = t a n (𝑒 𝑥), 𝑓' (𝑥) =$ $(A) s e c (𝑒 𝑥) t a n (𝑒 𝑥) \cdot 𝑒 𝑥 (B) s e c (𝑒 𝑥) \cdot 𝑒 𝑥 (C) s e c 2 (𝑒 𝑥)$ $(D) s e c 2 (𝑒 𝑥) \cdot 𝑒 𝑥 (E) n o n e o f t h e s e$

ABCDE

Chain Rule & Product Rule or Quotient Rule

The next few problems require using the Chain rule with the Product rule or with the Quotient rule.

Chain Rule Practice Problem #12

This problem combines the Product Rule with the Chain Rule.

Given $𝑓 (𝑥) = (𝑥 2 + 1) 7 (3 𝑥 - 7) 4, 𝑓' (𝑥) =$

(A) $28 (𝑥 2 + 1) 6 (3 𝑥 - 7) 3$

(B) $28 (𝑥 2 + 1) 6 (3 𝑥 - 7) 3 \cdot (2 𝑥) \cdot (3)$

(D) $7 (𝑥 2 + 1) 6 (3 𝑥 - 7) 4 + 4 (𝑥 2 + 1) 7 (3 𝑥 - 7) 3$

(E) none of these

ABCDE

Chain Rule Practice Problem #13

This problem combines the Quotient rule with the Chain rule.

Given

ℎ (𝑥) = 𝑒 2 𝑥 1 - 𝑥 2, 𝑓' (𝑥) =

(A) $(𝑒 2 𝑥 \cdot 2) (1 - 𝑥 2) - (𝑒 2 𝑥) (- 2 𝑥) (1 - 𝑥 2) 2 (- 2 𝑥)$

(B) $(𝑒 2 𝑥 \cdot 2) (1 - 𝑥 2) - (𝑒 2 𝑥) (- 2 𝑥) (1 - 𝑥 2) 2$

(D) $(𝑒 2 𝑥 \cdot 2) (1 - 𝑥 2) + (𝑒 2 𝑥) (- 2 𝑥) (1 - 𝑥 2) 2$

(E) none of these

ABCDE

Using the Chain Rule Multiple Times

The next few problems require using the Chain rule multiple times.

Chain Rule Practice Problem #14

This problem requires using the Chain Rule twice.

Differentiate $𝑓 (𝑥) = c o s (t a n (3 𝑥)) .$

Chain Rule Practice Problem #15

This problem requires using the Chain Rule twice.

Differentiate $𝑓 (𝑥) = \sqrt s i n (5 𝑥) .$

Chain Rule Practice Problem #16

This problem requires using the Chain Rule three times.

Differentiate

𝑓 (𝑥) = (1 + s i n 9 (2 𝑥 + 3)) 2 .

Hint: Recall that $s i n 9 (\dots) = [s i n (\dots)] 9 .$

Chain Rule: Problems and Solutions

Chain Rule

Chain Rule & Product Rule or Quotient Rule

Using the Chain Rule Multiple Times

More Problems and University Exam Problems