The Chain Rule from Tables and Graphs

In step 2 the outside derivative is evaluated at

g(a)

, the inside value. In a table, that usually means reading a different row from the one

a

is in, and reading the row of

a

instead is the usual slip.

From a table

Examples 1, 2, 3, and 5 use this table of the differentiable functions

f

and

g

$x$	$f(x)$	$f'(x)$	$g(x)$	$g'(x)$
$1$	$3$	$-1$	$2$	$4$
$2$	$4$	$2$	$3$	$-2$
$3$	$1$	$5$	$4$	$\tfrac{1}{2}$
$4$	$2$	$-3$	$1$	$5$

Example 1

Let

h(x) = f(g(x))

k(x) = g(f(x))

, and

q(x) = f\big(2g(x)\big)

. (a) Find the tangent line to the graph of

h

x = 1

. (b) Find

k'(3)

. (c) Find

q'(1)

(a) The inside value is

g(1) = 2

, so

h(1) = f(2) = 4

, and

\begin{aligned} h'(1) &= f'(g(1)) \cdot g'(1) \\ &= f'(2) \cdot 4 = 2 \cdot 4 = 8. \end{aligned}

The tangent line is

y = 4 + 8(x - 1)

(b) Now

f

is inside. Since

f(3) = 1

\begin{aligned} k'(3) &= g'(f(3)) \cdot f'(3) \\ &= g'(1) \cdot 5 = 4 \cdot 5 = 20. \end{aligned}

(c) The inside is

2g(x)

. At

x = 1

its value is

2g(1) = 4

, so the outside derivative comes from row

4

. The inside derivative is

2g'(1) = 8

\begin{aligned} q'(1) &= f'\big(2g(1)\big) \cdot 2g'(1) \\ &= f'(4) \cdot 8 = (-3)(8) = -24. \end{aligned}

Compare part (a). Doubling the inside changes the row you read and doubles the inside derivative.

Example 2

Let

m(x) = f\left(x^2\right)

p(x) = \big(g(x)\big)^3

, and

r(x) = f\left(g\left(x^2\right)\right)

. Find

m'(2)

p'(3)

, and

r'(2)

$m'(2)$ : the inside is

x^2

, so

\begin{aligned} m'(x) &= f'\left(x^2\right) \cdot 2x, \\ m'(2) &= f'(4) \cdot 4 = (-3)(4) = -12. \end{aligned}

$p'(3)$ : the outside is cubing, so

\begin{aligned} p'(3) &= 3\big(g(3)\big)^2 g'(3) \\ &= 3 \cdot 16 \cdot \tfrac{1}{2} = 24. \end{aligned}

$r'(2)$ : three layers, one factor each, working inward. At

x = 2

x^2 = 4

and

g(4) = 1

\begin{aligned} r'(2) &= f'\big(g(4)\big) \cdot g'(4) \cdot 2(2) \\ &= f'(1) \cdot 5 \cdot 4 \\ &= (-1)(5)(4) = -20. \end{aligned}

From a graph

Example 4

The graph of

f

is made of line segments. Let

w(x) = f(2x)

for

0 \le x \le 3

, and

v(x) = e^{f(x)}

The graph of f: segments joining (0, 1), (2, 3), (5, 0), and (6, 2).

(a) Find

w'\left(\frac{1}{2}\right)

and

w'(2)

. (b) Explain why

w'(1)

doesn't exist. (c) Find

v'(4)

(a) By the chain rule,

w'(x) = 2f'(2x)

. At

x = \frac{1}{2}

, the inside is

1

, on the first segment, with slope

1

, so

w'\left(\frac{1}{2}\right) = 2

. At

x = 2

, the inside is

4

, on the second segment, with slope

-1

, so

w'(2) = -2

(b) At

x = 1

, the inside is

2

, where

f

has a corner. For

x

just left of

1

2x

is just left of

2

, on the first segment, and for

x

just right of

1

it's on the second. So the one-sided derivatives of

w

1

are

2 \cdot 1 = 2

and

2 \cdot (-1) = -2

, and

w'(1)

doesn't exist. The graph of

w

is the graph of

f

squeezed toward the

y

-axis, and the corner comes along.

The graph of w(x) = f(2x): the same shape at half the width. Every slope is doubled, and the corners are now at x = 1 and x = 2.5.

(c)

v'(x) = e^{f(x)}f'(x)

. From the graph,

f(4) = 1

and

f'(4) = -1

, so

v'(4) = e^1 \cdot (-1) = -e

From a table

From a graph

A product with a composite factor