Use the chain rule with a table

Builds on applying the chain rule completely. 6 minutes.

Reading one row, then another

The table lists \(f\), \(f'\), \(g\) and \(g'\) at four inputs. Let \(h(x) = f(g(x))\). The goal is \(h'(2)\).

Find the row \(x = 2\). It shows \(g(2) = 4\) and \(g'(2) = -3\).

The value \(g(2) = 4\) is the input of \(f\), so the orange arrow carries it from the row \(x = 2\) to the row \(x = 4\).

In the row \(x = 4\), the table shows \(f'(4) = -1\), so the chain rule gives \(h'(2) = f'(4)\cdot g'(2) = (-1)(-3) = 3\).

Remove the numbers. Each row is named by its input, and the row for \(a\) holds the inner value \(g(a)\) and the inner slope \(g'(a)\).

The inner value names a second row, and the arrow goes there. That row holds the outer slope \(f'(g(a))\).

For any table, the inner slope comes from the starting row, and the outer slope comes from the row named by the inner value.

Start at the input \(a\), which is 2 in the table. In its row, read the inner value and the inner slope.
\[\cy{g(a)} \qquad \cy{g'(a)}\]
Go to the row named by \(g(a)\). Read the outer slope there.
\[f'(\cy{g(a)})\]
The chain rule multiplies the outer slope by the inner slope.
\[h'(a) = f'(g(a)) \cdot g'(a)\]
The rule needs \(g\) to be differentiable at \(a\) and \(f\) to be differentiable at \(g(a)\).

Suppose \(g(2) = 5\). The arrow goes to the row \(x = 5\), but the table has no such row.

The table does not give \(f'(5)\), so it cannot give \(h'(2)\). A table gives \(h'(a)\) only if it lists the inner value \(g(a)\) as an input.

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

In the row \(x = 3\), read \(g(3) = 1\) and \(g'(3) = 2\).
The same row shows \(f'(3) = 7\), but \(h'(3)\) does not use it.
The inner value is 1, so the arrow goes to the row \(x = 1\), where \(f'(1) = -2\).
Multiply the outer slope by the inner slope.
\[\begin{aligned} h'(3) &= f'(g(3)) \cdot g'(3) \\ &= f'(1) \cdot 2 = \cy{-4} \end{aligned}\]

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

Here \(f\) is the inner function. In the row \(x = 2\), read \(f(2) = 1\) and \(f'(2) = 6\).
A tempting wrong move: The outer function is \(g\). Read \(g'\) in the same row, where \(g'(2) = -3\).
\[\begin{aligned} H'(2) &= g'(2) \cdot f'(2) \\ &= (-3)(6) = -18 \end{aligned}\]
Because \(f(2) = 1\), the orange arrow runs from the row \(x = 2\) to the row \(x = 1\), and the outer slope belongs to that row. The red cell, \(-3\), is in the starting row, so it is the wrong value for the outer slope.
Read \(g'\) in the row \(x = 1\), where \(g'(1) = 5\). Multiply.
\[\begin{aligned} H'(2) &= g'(f(2)) \cdot f'(2) \\ &= g'(1) \cdot 6 = \cy{30} \end{aligned}\]

Use the table. Let \(K(x) = f(f(x))\). Find \(K'(4)\).

Both layers are f, so the value f(4) names the row where you read the outer slope f′.

Solution

In the row \(x = 4\), read \(f(4) = 3\) and \(f'(4) = -1\).
Because \(f(4) = 3\), the arrow goes to the row \(x = 3\), where \(f'(3) = 7\).
Multiply the outer slope by the inner slope.
\[\begin{aligned} K'(4) &= f'(f(4)) \cdot f'(4) \\ &= 7 \cdot (-1) = -7 \end{aligned}\]

Let \(h(x) = f(g(x))\). The figure shows the table, with an orange outline on the row \(x = 1\).

You read the inner slope in the starting row, and the outer slope in the row named by the inner value.

This works because the outer slope belongs to the input of the outer function, and that input is the inner value, not x.

The same steps apply whenever a problem gives values of the functions and their derivatives in a table instead of formulas.

This is one of four skills in topic 3.1 of Which & Why: AP Calculus.