Wemazi

Topic 2.8

Find a missing value in the rule

Builds on applying a product derivative. 8 minutes.

Which rate makes the strips add up?

A rectangle has width \(f(a)=5\) and height \(g(a)=3\). The width grows at the rate \(f'(a)=2\), and the rate \(g'(a)\) of the height is unknown.

The rate of the area \(fg\) is the sum of two strips. The strip on the right is \(f'\) times \(g\), which is \(2\cdot3=6\), and it is green because the width grows. The strip on top is \(f\) times \(g'\), which is \(5\,g'\).

The rate of the area is \((fg)'(a)=1\). Move the slider for \(g'(a)\) until the sum of the strips is 1.

The sum is 1 only when the strip on top is \(1-6=-5\), so \(5\,g'(a)=-5\) and \(g'(a)=-1\). This strip is red because the height shrinks.

One equation with one unknown

The product rule is an equation that links the rate of the product to four numbers: \(f(a)\), \(g(a)\), \(f'(a)\) and \(g'(a)\). If you know the rate and three of the numbers, you find the fourth by solving the equation.

To find \(g'(a)\), subtract the known strip from the rate. What remains is the strip on top, so divide it by its width \(f(a)\).

Move the slider for the rate of the product. The strip on top changes so that the two strips always add up to the rate.

Notation

  1. The rate of the product is the strip on the right plus the strip on top.

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

  2. If \(g'(a)\) is the unknown, subtract the known strip. Then divide by \(f(a)\).

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

    The same method finds any other unknown. Isolate its strip. Then divide by the number that multiplies it.

  3. The division needs \(f(a)\ne0\), because you cannot divide by 0.

When it fails

When \(f(a)=0\), the strip on top has the width 0. The rule becomes \((fg)'(a)=f'(a)\,g(a)\), whatever \(g'(a)\) is.

Move \(f(a)\) to 0. Then move \(g'(a)\). The strip on top does not change the sum, so the rate of the product cannot tell you \(g'(a)\).

Example 1: a missing derivative

Let \(h(x)=x^2 g(x)\). Suppose \(g(3)=2\) and \(h'(3)=30\). Find \(g'(3)\).

  1. At \(x=3\), the first factor \(x^2\) is 9 and its derivative \(2x\) is 6. These are the widths of the rectangle and of the strip on the right.

  2. Write the product rule at \(x=3\), with \(g'(3)\) as the unknown.

    \[h'(3) = 6\cdot g(3) + 9\cdot g'(3)\]

  3. Substitute \(h'(3)=30\) and \(g(3)=2\). The strip on the right is \(6\cdot2=12\).

    \[30 = 12 + 9\,g'(3)\]

  4. Subtract 12. Then divide by 9.

    \[g'(3) = \frac{30-12}{9} = 2\]

  5. To check, the strips are 12 and \(9\cdot2=18\), and their sum is 30.

Example 2: a missing value

Let \(h(x)=f(x)\,g(x)\). Suppose \(f'(2)=3\), \(g(2)=2\) and \(g'(2)=-1\). The graph of \(h\) has a horizontal tangent at \(x=2\). Find \(f(2)\).

  1. A horizontal tangent has slope 0, so \(h'(2)=0\) and the two strips must cancel. The strip on top is red because \(g'(2)\) is negative, and the slider sets its width \(f(2)\).

  2. A tempting wrong move: A product is 0 only when a factor is 0, so \(f(2)=0\).

    \[\cr{f(2)=0}\]

    The 0 is \(h'(2)\), not \(h(2)\), and the product rule makes \(h'(2)\) a sum of two strips. With \(f(2)=0\), the red strip on top disappears and the sum is the green strip, \(3\cdot2=6\), not 0.

  3. Write the product rule at \(x=2\), with \(f(2)\) as the unknown.

    \[0 = 3\cdot2 + f(2)\cdot(-1)\]

  4. Solve the equation.

    \[6 - f(2) = 0 \;\Rightarrow\; f(2) = 6\]

  5. To check, the green strip is 6 and the red strip is 6 · (−1) = −6, so they cancel.

Practice

Let \(p(x)=x^3 g(x)\). Suppose \(g(2)=1\) and \(p'(2)=4\). Find \(g'(2)\).

At \(x=2\), \(x^3\) is 8 and its derivative \(3x^2\) is 12, so the strip on the right is \(12\cdot1=12\). The strip on top must make the sum 4.

Solution

  1. Write the product rule at \(x=2\). The derivative of \(x^3\) is \(3x^2\), which is 12 at \(x=2\).

    \[p'(2) = 12\cdot g(2) + 8\cdot g'(2)\]

  2. Substitute \(p'(2)=4\) and \(g(2)=1\).

    \[4 = 12 + 8\,g'(2)\]

  3. Subtract 12. Then divide by 8. The result is negative, so g decreases at x = 2 and the strip on top is red.

    \[g'(2) = \frac{4-12}{8} = -1\]

Questions

The figure shows a rectangle for the product \(fg\) at one input \(a\).

  1. The width is \(f(a)=4\), and the height is \(g(a)=3\). The total rate is \((fg)'(a)=14\), and the green strip on the right is 6. What is \(g'(a)\)?

    • 3.5
    • 8
    • 2
    Answer

    2

  2. Now \(f(a)=0\), \(f'(a)=2\), \(g(a)=3\) and \((fg)'(a)=6\). What does this tell you about \(g'(a)\)?

    • \(g'(a)=0\)
    • \(g'(a)=2\)
    • Nothing: \(g'(a)\) can have any value
    Answer

    Nothing: \(g'(a)\) can have any value

  3. Let \(h(x)=(x+1)\,g(x)\). Suppose \(g(3)=2\) and \(h'(3)=11\). What is \(g'(3)\)?

    • 9
    • \(\frac{13}{4}\)
    • \(\frac{9}{4}\)
    • \(\frac{11}{4}\)
    Answer

    \(\frac{9}{4}\)

Summary

You find a missing value by writing the product rule at the input, substituting what you know, and solving for the unknown.

This works because the rule is an equation, and it has one solution when the number that multiplies the unknown is not 0.

You use it when a problem gives the derivative of a product and asks for a value or a derivative of one factor.

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