Wemazi

Topic 2.8

Use a product rule with data

Builds on differentiating a product. 7 minutes.

Four numbers from one column

The table gives the values of two functions \(f\) and \(g\), and of their derivatives \(f'\) and \(g'\), at four inputs.

Choose an input with the buttons. Its column lights up in orange, and the working for it appears below the table.

The figure starts at \(x=2\), where the product rule needs four numbers from the column: \(f(2)=3\), \(g(2)=4\), \(f'(2)=-1\) and \(g'(2)=6\).

The first term is \(f'\) times \(g\), which is \((-1)(4)=-4\). The second term is \(f\) times \(g'\), which is \((3)(6)=18\). The derivative of the product is \((fg)'(2)=-4+18=14\).

Choose another column. The answer changes, because all four numbers change with the input.

One input at a time

The table now holds symbols. The input is \(a\), and its column holds \(f(a)\), \(f'(a)\), \(g(a)\) and \(g'(a)\).

The first term of the rule takes \(f'(a)\) and \(g(a)\), and the second term takes \(f(a)\) and \(g'(a)\). All four numbers come from column \(a\).

Press Read g′ from the other column. The sum now uses \(g'(b)\) and turns red, because it mixes two inputs and so is not the derivative of the product at any input.

Notation

  1. At the input \(a\), the derivative of the product \(fg\) is the sum of two terms. Each term has one derivative and one value.

    \[(fg)'(a) = \cy{f'(a)}\,g(a) + f(a)\,\cy{g'(a)}\]

  2. The rule needs \(f\) and \(g\) to be differentiable at \(a\). A table that gives \(f'(a)\) and \(g'(a)\) as numbers shows that they are.

When it fails

Press Give f a corner at x = 4. The graph of \(f\) now has a corner at \(x=4\), so \(f'(4)\) does not exist, and its cell reads none in red.

The rule needs the number \(f'(4)\), so it gives no value for \((fg)'(4)\). A question about the product at \(x=4\) needs another method.

Example 1: one factor is x

Let \(h(x)=x\,g(x)\). Two values are known: \(g(5)=2\) and \(g'(5)=-3\). Find \(h'(5)\).

  1. The first factor is \(u(x)=x\). At \(x=5\) its value is 5, and its derivative is 1.

  2. Apply the rule: the derivative of the first factor times \(g\), plus the first factor times \(g'\).

    \[h'(5) = 1\cdot g(5) + 5\cdot g'(5)\]

  3. Substitute \(g(5)=2\) and \(g'(5)=-3\).

    \[h'(5) = 1\cdot 2 + 5\cdot(-3) = -13\]

Example 2: values from tangent lines

At \(x=2\), the tangent line to \(f\) is \(y=3x-2\). The tangent line to \(g\) is \(y=10-2x\). Find \((fg)'(2)\).

  1. A tangent line touches the curve at \(x=2\). At that point the height of the line is the value of the function, and the slope of the line is the derivative.

  2. A tempting wrong move: Read \(f(2)\) from the y-intercept of the tangent line.

    \[\cr{f(2)=-2}\]

    The tangent line touches the curve at \(x=2\), where its height is \(3\cdot2-2=4\). The number \(-2\) is the height of the line at \(x=0\), a different input.

  3. Substitute \(x=2\) into each line to find the values. The slopes of the lines are the derivatives.

    \[\begin{aligned}f(2)=4,\; f'(2)=3; \\ g(2)=6,\; g'(2)=-2\end{aligned}\]

  4. Apply the rule with these four numbers.

    \[(fg)'(2) = 3\cdot 6 + 4\cdot(-2) = 10\]

Practice

Use the table of the opening. Let \(k(x)=x^2 f(x)\). Find \(k'(3)\).

Find the column \(x=3\) in the table. The first factor \(x^2\) is not in the table, so find its value and its derivative at \(x=3\) yourself.

Solution

  1. The first factor is \(x^2\). At \(x=3\) its value is 9, and its derivative \(2x\) is 6. The column \(x=3\) gives \(f(3)=-2\) and \(f'(3)=4\).

  2. Apply the rule.

    \[k'(3) = 6\cdot f(3) + 9\cdot f'(3)\]

  3. Substitute the values.

    \[k'(3) = 6(-2) + 9(4) = 24\]

Questions

Questions 1 and 2 use the table of the opening. Question 3 needs a calculation.

  1. Use the column \(x=1\). What is \((fg)'(1)\)?

    • −6
    • 10
    • 11
    • 19
    Answer

    11

  2. Use the column \(x=4\). What is \((fg)'(4)\)?

    • 4
    • −8
    • −4
    Answer

    −4

  3. Let \(h(x)=x^2 g(x)\), with \(g(2)=5\) and \(g'(2)=-1\). What is \(h'(2)\)?

    • 16
    • −4
    • 24
    • 20
    Answer

    16

Summary

You find the derivative of a product at one input by adding f′g and f g′, with all four numbers read from that input.

This works because the product rule holds at each single input, so every number in it must belong to that input.

You use it when a table, a graph or tangent lines give values and slopes instead of formulas.

This is one of four skills in topic 2.8 of Which & Why: AP Calculus.