Use combined rules with a table
Four numbers from one row
The table gives \(f\), \(g\) and their derivatives at three inputs, and the row for \(x = 2\) is outlined.
For the product \(p = f\cdot g\), the rule needs four numbers, \(f(2)\), \(f'(2)\), \(g(2)\) and \(g'(2)\). The figure marks these cells in orange, and they give \(p'(2) = 3\cdot 1 + 5\cdot(-2) = -7\).
For the composite \(c(x) = f(g(x))\), the inside value \(g(2) = 1\) is the input of \(f\), so the arrow carries it to row 1, where you read \(f'(1) = -1\). The result is \(c'(2) = f'(1)\cdot g'(2) = (-1)(-2) = 2\).
The product rule reads one row, but the chain rule reads two.
The row each rule reads
The table keeps its layout without the numbers, and each row is named by its input.
The product rule and the quotient rule at \(a\) read all four numbers from row \(a\).
The chain rule at \(a\) reads \(g(a)\) and \(g'(a)\) in row \(a\). The arrow then carries \(g(a)\) to the row it names, where \(f'(g(a))\) is read.
Every number must be read at the input that the rule asks for.
Notation
For a product, read all four values at the same input \(a\). In the table, \(a\) is 2.
\[\cy{(fg)'(a)} = f'(a)\,g(a) + f(a)\,g'(a)\]For a quotient, also read at \(a\). The bottom \(g(a)\) must not be 0, because the rule divides by \(g(a)^2\).
\[\begin{aligned} &\cy{\left(\dfrac fg\right)'(a)} \\ &\quad = \dfrac{f'(a)\,g(a) - f(a)\,g'(a)}{g(a)^2} \end{aligned}\]For a composition, read \(g(a)\) and \(g'(a)\) at \(a\). Then read \(f'\) at the input \(g(a)\), which is row 1 here because \(g(2) = 1\).
\[\cy{\big(f(g(x))\big)'(a)} = f'\big(g(a)\big)\,g'(a)\]The table must contain every input that the rule names. If it does not, the table cannot give the answer.
When it fails
Suppose \(g(2)\) were 4, shown in red. The chain rule then needs \(f'(4)\), but the table has no row for 4. The red question mark stands where \(f'(4)\) would be.
The product at \(x = 2\) still works, because it reads only row 2. Only the chain rule leaves the row you start in.
Example 1: a quotient
Let \(k(x) = \dfrac{f(x)}{g(x)}\). Use the table to find \(k'(3)\).
The last operation is a division, so the quotient rule applies at \(x = 3\). All four numbers come from row 3: \(f(3) = 2\), \(f'(3) = 6\), \(g(3) = 2\) and \(g'(3) = 5\).
Put the numbers into the rule.
\[k'(3) = \dfrac{6\cdot 2 - 2\cdot 5}{2^2}\]The top is 12 − 10 = 2, and the bottom is 4.
\[k'(3) = \dfrac{1}{2}\]
Example 2: a product with a composition
Let \(m(x) = x^2\,f(g(x))\). Use the table to find \(m'(1)\).
The last operation is a product, so the product rule comes first. The chain rule gives the slope of the second factor.
\[\begin{aligned} m'(x) &= 2x\,f(g(x)) \\ &\quad + x^2\,f'(g(x))\,g'(x) \end{aligned}\]Read \(g\) in row 1: \(g(1) = 3\) and \(g'(1) = 2\). The value 3 is the input of \(f\), so \(f(g(1)) = f(3) = 2\).
A tempting wrong move: Read \(f'\) in row 1, where \(x = 1\).
\[\begin{aligned} m'(1) &= 2\cdot 1\cdot 2 + 1\cdot(-1)\cdot 2 \\ &= 2 \end{aligned}\]The arrow shows that \(f\) is evaluated at 3, not at 1, so the slope of \(f\) is read in row 3. The red cell \(f'(1) = -1\) is in the wrong row.
Read \(f'\) at the input \(g(1) = 3\), which gives \(f'(3) = 6\).
\[m'(1) = 2\cdot 1\cdot 2 + 1\cdot 6\cdot 2 = 16\]
Practice
Let \(P(x) = x\,g(f(x))\). Use the table to find \(P'(3)\).
Row 3 gives f(3), and that value names the row where you read g and g′.
Solution
The last operation is a product, so the product rule comes first. The chain rule gives the slope of the second factor.
\[\begin{aligned} P'(x) &= g(f(x)) \\ &\quad + x\,g'(f(x))\,f'(x) \end{aligned}\]Read \(f\) in row 3: \(f(3) = 2\) and \(f'(3) = 6\). The value 2 is the input of \(g\), so the arrow points to row 2.
Read \(g\) in row 2: \(g(2) = 1\) and \(g'(2) = -2\). The first term is \(g(f(3)) = g(2) = 1\).
\[P'(3) = 1 + 3\cdot(-2)\cdot 6 = -35\]
Questions
The table gives \(f\), \(g\) and their derivatives. After you answer, the figure marks the cells that the question needs.
What is \((fg)'(3)\)?
- 22
- 30
- 2
- 12
Answer
22
What is the derivative of \(f(g(x))\) at \(x = 3\)?
- 15
- 30
- −6
- 3
Answer
15
Let \(H(x) = \big(f(x)\big)^3\). What is \(H'(2)\)?
- 225
- 75
- 375
- 125
Answer
225
Summary
You read each value that a rule needs from the row it names, then put the values into the rule.
This works because a rule at one input uses only numbers from that input, except the chain rule, which reads \(f'\) in row \(g(a)\).
You use it when a table gives f, g and their derivatives and asks for the derivative of a combination.
This is one of four skills in topic 3.5 of Which & Why: AP Calculus.