Chain Rule from Tables and Graphs

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Key Ideas Easy Medium Hard

\((f\circ g)'(x) = f'\!\bigl(g(x)\bigr)\cdot g'(x)\)

What to Remember

Read the inside first

\((f\circ g)'(x) = f'\!\bigl(g(x)\bigr)\cdot g'(x)\)

At a number \(a\), find \(g(a)\) in the table. Use that output as the input of \(f'\). Then multiply by \(g'(a)\).

The order of the composition matters

\((g\circ f)'(a) = g'\!\bigl(f(a)\bigr)\cdot f'(a)\)

Swap the functions and you are answering a different question.

A formula wrapped around a table function

For \(h(x) = f(x^2)\), the inside derivative is \(2x\). Multiply \(f'(x^2)\) by \(2x\).

A corner blocks the chain rule

On the graph below, \(f\) has a corner at \(x = 1\), so \(f'(1)\) does not exist. If \(g(a) = 1\) and \(g'(a)\) is not 0, then \((f\circ g)'(a)\) does not exist.

Level 1

Easy

The table gives every value you need. Compose, then multiply.

Example

Use the table. Find \((f\circ g)'(2)\).

\(x\)	\(f(x)\)	\(f'(x)\)	\(g(x)\)	\(g'(x)\)
1	4	\(-2\)	3	5
2	3	6	1	\(-3\)
3	\(-1\)	2	2	4
4	2	1	\(-1\)	2

\((f\circ g)'(x) = f'\!\bigl(g(x)\bigr)\cdot g'(x)\)

\(g(2) = 1\)

\(f'\!\bigl(g(2)\bigr) = f'(1) = -2\)

\(g'(2) = -3\)

\((-2)(-3)\)

\(6\)

Answer: \(6\)

Use the table. Find \((f\circ g)'(1)\).

\(x\)	\(f(x)\)	\(f'(x)\)	\(g(x)\)	\(g'(x)\)
1	4	\(-2\)	3	5
2	3	6	1	\(-3\)
3	\(-1\)	2	2	4
4	2	1	\(-1\)	2

Show Answer

\((f\circ g)'(x) = f'\!\bigl(g(x)\bigr)\cdot g'(x)\)

\(g(1) = 3\)

\(f'(3) = 2\)

\(g'(1) = 5\)

\(2\cdot 5\)

\(10\)

Answer: \(10\)

Use the table. Find \((g\circ f)'(1)\).

\(x\)	\(f(x)\)	\(f'(x)\)	\(g(x)\)	\(g'(x)\)
1	4	\(-2\)	3	5
2	3	6	1	\(-3\)
3	\(-1\)	2	2	4
4	2	1	\(-1\)	2

Show Answer

\((g\circ f)'(x) = g'\!\bigl(f(x)\bigr)\cdot f'(x)\)

\(f(1) = 4\)

\(g'(4) = 2\)

\(f'(1) = -2\)

\(2\cdot(-2)\)

\(-4\)

Answer: \(-4\)

Use the table. Find \((f\circ g)'(3)\).

\(x\)	\(f(x)\)	\(f'(x)\)	\(g(x)\)	\(g'(x)\)
1	4	\(-2\)	3	5
2	3	6	1	\(-3\)
3	\(-1\)	2	2	4
4	2	1	\(-1\)	2

Show Answer

\((f\circ g)'(x) = f'\!\bigl(g(x)\bigr)\cdot g'(x)\)

\(g(3) = 2\)

\(f'(2) = 6\)

\(g'(3) = 4\)

\(6\cdot 4\)

\(24\)

Answer: \(24\)

Use the table. Find \((f\circ f)'(2)\).

\(x\)	\(f(x)\)	\(f'(x)\)	\(g(x)\)	\(g'(x)\)
1	4	\(-2\)	3	5
2	3	6	1	\(-3\)
3	\(-1\)	2	2	4
4	2	1	\(-1\)	2

Show Answer

\((f\circ f)'(x) = f'\!\bigl(f(x)\bigr)\cdot f'(x)\)

\(f(2) = 3\)

\(f'(3) = 2\)

\(f'(2) = 6\)

\(2\cdot 6\)

\(12\)

Answer: \(12\)

Use the table. Find \((g\circ g)'(1)\).

\(x\)	\(f(x)\)	\(f'(x)\)	\(g(x)\)	\(g'(x)\)
1	4	\(-2\)	3	5
2	3	6	1	\(-3\)
3	\(-1\)	2	2	4
4	2	1	\(-1\)	2

Show Answer

\((g\circ g)'(x) = g'\!\bigl(g(x)\bigr)\cdot g'(x)\)

\(g(1) = 3\)

\(g'(3) = 4\)

\(g'(1) = 5\)

\(4\cdot 5\)

\(20\)

Answer: \(20\)

Level 2

Medium

The chain rule still applies when the inside is a formula and the outside comes from the table.

Example

Use the table. Let \(h(x) = f(x^2 + 1)\). Find \(h'(1)\).

\(x\)	\(f(x)\)	\(f'(x)\)	\(g(x)\)	\(g'(x)\)
1	4	\(-2\)	3	5
2	3	6	1	\(-3\)
3	\(-1\)	2	2	4
4	2	1	\(-1\)	2

\(h'(x) = f'(x^2 + 1)\cdot 2x\)

Substitute \(x = 1\).

\(x^2 + 1 = 2\)

\(f'(2) = 6\)

\(2x = 2\)

\(6\cdot 2\)

\(12\)

Answer: \(12\)

Use the table. Let \(h(x) = g(3x)\). Find \(h'(1)\).

\(x\)	\(f(x)\)	\(f'(x)\)	\(g(x)\)	\(g'(x)\)
1	4	\(-2\)	3	5
2	3	6	1	\(-3\)
3	\(-1\)	2	2	4
4	2	1	\(-1\)	2

Show Answer

\(h'(x) = g'(3x)\cdot 3\)

Substitute \(x = 1\).

\(g'(3) = 4\)

\(4\cdot 3\)

\(12\)

Answer: \(12\)

Use the table. Let \(k(x) = f(2x)\). Find \(k'(1)\).

\(x\)	\(f(x)\)	\(f'(x)\)	\(g(x)\)	\(g'(x)\)
1	4	\(-2\)	3	5
2	3	6	1	\(-3\)
3	\(-1\)	2	2	4
4	2	1	\(-1\)	2

Show Answer

\(k'(x) = f'(2x)\cdot 2\)

\(f'(2) = 6\)

\(6\cdot 2\)

\(12\)

Answer: \(12\)

Use the table. Let \(p(x) = \bigl(g(x)\bigr)^2\). Find \(p'(2)\).

\(x\)	\(f(x)\)	\(f'(x)\)	\(g(x)\)	\(g'(x)\)
1	4	\(-2\)	3	5
2	3	6	1	\(-3\)
3	\(-1\)	2	2	4
4	2	1	\(-1\)	2

Show Answer

\(p'(x) = 2\, g(x)\cdot g'(x)\)

\(g(2) = 1\)

\(g'(2) = -3\)

\(2(1)(-3)\)

\(-6\)

Answer: \(-6\)

Use the table. Which value equals \((f\circ g)'(2)\)?

\(x\)	\(f(x)\)	\(f'(x)\)	\(g(x)\)	\(g'(x)\)
1	4	\(-2\)	3	5
2	3	6	1	\(-3\)
3	\(-1\)	2	2	4
4	2	1	\(-1\)	2

A. \(6\)

B. \(-6\)

C. \(-5\)

D. \(10\)

Show Answer

\((f\circ g)'(x) = f'\!\bigl(g(x)\bigr)\cdot g'(x)\)

\(f'\!\bigl(g(2)\bigr) = f'(1) = -2\)

\(g'(2) = -3\)

\((-2)(-3) = 6\)

Answer: A. \(6\)

Use the graph. The slope is 2 on the open interval \((-1, 1)\) and the slope is \(-2\) on \((1, 4)\). Let \(g(1) = 0\) and \(g'(1) = 3\). Find \((f\circ g)'(1)\).

Graph of f

Show Answer

\((f\circ g)'(x) = f'\!\bigl(g(x)\bigr)\cdot g'(x)\)

\(g(1) = 0\)

On \((-1, 1)\), the slope of \(f\) is 2, so \(f'(0) = 2\).

\(g'(1) = 3\)

\(2\cdot 3\)

\(6\)

Answer: \(6\)

Level 3

Hard

A second composition, a quotient, or a corner.

Example

Use the table. Let \(h(x) = f\!\bigl(g(x^2)\bigr)\). Find \(h'(1)\).

\(x\)	\(f(x)\)	\(f'(x)\)	\(g(x)\)	\(g'(x)\)
1	4	\(-2\)	3	5
2	3	6	1	\(-3\)
3	\(-1\)	2	2	4
4	2	1	\(-1\)	2

\(h'(x) = f'\!\bigl(g(x^2)\bigr)\cdot g'(x^2)\cdot 2x\)

Substitute \(x = 1\).

\(g(1) = 3\)

\(f'(3) = 2\)

\(g'(1) = 5\)

\(2x = 2\)

\(2\cdot 5\cdot 2\)

\(20\)

Answer: \(20\)

Use the table. Let \(q(x) = g\!\bigl(f(x)\bigr) + x\). Find \(q'(2)\).

\(x\)	\(f(x)\)	\(f'(x)\)	\(g(x)\)	\(g'(x)\)
1	4	\(-2\)	3	5
2	3	6	1	\(-3\)
3	\(-1\)	2	2	4
4	2	1	\(-1\)	2

Show Answer

\(q'(x) = g'\!\bigl(f(x)\bigr)\cdot f'(x) + 1\)

\(f(2) = 3\)

\(g'(3) = 4\)

\(f'(2) = 6\)

\(4\cdot 6 + 1\)

\(25\)

Answer: \(25\)

Use the table. Let \(s(x) = \dfrac{f\!\bigl(g(x)\bigr)}{x}\). Find \(s'(2)\).

\(x\)	\(f(x)\)	\(f'(x)\)	\(g(x)\)	\(g'(x)\)
1	4	\(-2\)	3	5
2	3	6	1	\(-3\)
3	\(-1\)	2	2	4
4	2	1	\(-1\)	2

Show Answer

\(s'(x) = \dfrac{(f\circ g)'(x)\cdot x - f\!\bigl(g(x)\bigr)\cdot 1}{x^2}\)

\((f\circ g)'(2) = f'(1)\cdot g'(2) = (-2)(-3) = 6\)

\(f\!\bigl(g(2)\bigr) = f(1) = 4\)

\(\dfrac{6\cdot 2 - 4\cdot 1}{2^2}\)

\(\dfrac{8}{4}\)

\(2\)

Answer: \(2\)

Use the graph and the values \(g(2) = 1\), \(g'(2) = 4\). Does \((f\circ g)'(2)\) exist?

Graph of f

Show Answer

The graph of \(f\) has a corner at \(x = 1\).

\(f'(1)\) does not exist.

\(g(2) = 1\) and \(g'(2) = 4 \neq 0\).

The chain rule needs \(f'\) at \(g(2)\).

\((f\circ g)'(2)\) does not exist.

Answer: does not exist

Use the table. Let \(k(x) = f(2x)\). Find \(k'(2)\).

\(x\)	\(f(x)\)	\(f'(x)\)	\(g(x)\)	\(g'(x)\)
1	4	\(-2\)	3	5
2	3	6	1	\(-3\)
3	\(-1\)	2	2	4
4	2	1	\(-1\)	2

Show Answer

\(k'(x) = f'(2x)\cdot 2\)

Substitute \(x = 2\).

\(f'(4) = 1\)

\(1\cdot 2\)

\(2\)

Answer: \(2\)

Use the table. Which value equals \((g\circ g)'(3)\)?

\(x\)	\(f(x)\)	\(f'(x)\)	\(g(x)\)	\(g'(x)\)
1	4	\(-2\)	3	5
2	3	6	1	\(-3\)
3	\(-1\)	2	2	4
4	2	1	\(-1\)	2

A. \(-12\)

B. \(12\)

C. \(-3\)

D. \(8\)

Show Answer

\((g\circ g)'(3) = g'\!\bigl(g(3)\bigr)\cdot g'(3)\)

\(g(3) = 2\)

\(g'(2) = -3\)

\(g'(3) = 4\)

\((-3)(4) = -12\)

Answer: A. \(-12\)

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