Wemazi

Topic 2.9

Use a quotient rule with data

Builds on differentiating a quotient. 7 minutes.

Reading one column of a table

A table gives the values of \(f\) and \(g\) and their slopes \(f'\) and \(g'\) at each input \(x\), but no formulas. The quotient rule needs only these four numbers to find the slope of \(Q = f/g\).

The chosen column, \(x = 1\), is orange. It gives \(f = 5\), \(f' = 2\), \(g = 2\) and \(g' = 1\), and the working below the table puts each number into the rule.

The top of the rule is \(f'g - fg' = 2\cdot 2 - 5\cdot 1 = -1\), and the bottom is \(g^2 = 4\), so \(Q'(1) = -\tfrac14\).

Choose the column \(x = 2\). The rule is the same, and only the four numbers change, so \(Q'(2) = \frac{(-4)(4) - (3)(3)}{4^2} = -\frac{25}{16}\).

Four numbers at one input

The inputs and the numbers can be anything. At an input \(a\), the rule uses four numbers:\(f(a)\), \(f'(a)\), \(g(a)\) and \(g'(a)\).

Each button lights one column, and the rule uses only that column, because entries from two columns belong to two different inputs.

In every column, \(f\) is the top and \(g\) is the bottom, and the bottom of the rule is \(g(a)^2\), which is not negative.

Notation

  1. A column gives, for one input \(a\), the values \(f(a)\) and \(g(a)\) and the slopes \(f'(a)\) and \(g'(a)\).

    \[f(a),\quad f'(a),\quad g(a),\quad g'(a)\]

  2. Test the bottom first. If \(g(a) = 0\), the ratio has no value at \(a\), so it has no slope there. Here \(g(2) = 4\).

    \[g(a) \ne 0\]

  3. Substitute the four numbers into the rule, which needs no formula for \(f\) or \(g\).

    \[Q'(a) = \frac{\cy{f'(a)}\,\cy{g(a)} - \cy{f(a)}\,\cy{g'(a)}}{\cy{g(a)}^2}\]

  4. The rule depends on which function is on top, not on its letter. For \(R = g/f\), \(g\) is the top and \(f\) is the bottom, so \(R'(2) = \tfrac{25}{9}\).

    \[\begin{aligned}\left(\frac{g}{f}\right)' \\ &= \frac{g'f - gf'}{f^2}\end{aligned}\]

When it fails

In the column \(x = 4\), \(g(4) = 0\). The ratio \(Q = f/g\) has no value at 4, because division by 0 is not possible.

The column still lists \(f' = 7\) and \(g' = 6\), but the bottom of the rule is \(0^2 = 0\), so \(Q'(4)\) does not exist.

Example 1: a negative bottom

A table gives \(f(2) = 7\), \(f'(2) = -2\), \(g(2) = -1\) and \(g'(2) = 3\). Find \(h'(2)\) for \(h = f/g\).

  1. The bottom is \(g(2) = -1\), which is not 0, so the rule applies.

  2. Find the two products in the top of the rule. The first is \(f'g = (-2)(-1) = 2\), and the second is \(fg' = 7\cdot 3 = 21\).

    \[f'g = 2,\qquad fg' = 21\]

  3. Subtract the second product from the first. Then divide by \(g^2 = (-1)^2 = 1\), where the square removes the minus sign.

    \[h'(2) = \frac{2 - 21}{1} = \cy{-19}\]

Example 2: swapping the roles

A table gives \(f(3) = 2\), \(f'(3) = 5\), \(g(3) = 6\) and \(g'(3) = -1\). Find the slope of \(R = g/f\) at \(x = 3\).

  1. The ratio \(R\) has \(g\) on top and \(f\) on the bottom, and the bottom \(f(3) = 2\) is not 0. The graph of \(R\) is red because \(R\) falls near \(x = 3\), and the orange line is its tangent at \(x = 3\).

  2. A tempting wrong move: Use the rule as it stands for \(f/g\).

    \[\begin{aligned}\frac{f'g - fg'}{g^2} &= \frac{5\cdot 6 - 2\cdot(-1)}{6^2} \\ &= \frac{8}{9}\end{aligned}\]

    This is the slope of \(f/g\), because it puts \(f'\) first and divides by \(g^2\). The red dashed line with this slope rises, but the orange tangent of \(R\) falls.

  3. For \(g/f\), put the derivative of the top, \(g'\), times the bottom \(f\) first.

    \[\begin{aligned}R'(3) &= \frac{g'f - gf'}{f^2} \\ &= \frac{(-1)(2) - (6)(5)}{2^2} = \cy{-8}\end{aligned}\]

  4. As a check, \(R = 1/Q\) with \(Q = f/g\), and at \(x = 3\), \(Q = \tfrac13\) and \(Q' = \tfrac89\). The quotient rule with top 1 gives \(R' = -Q'/Q^2 = -\tfrac89 \div \tfrac19 = -8\).

Practice

A table gives \(m(4) = 10\) and \(m'(4) = -6\). Find the slope of \(p(x) = \dfrac{m(x)}{x}\) at \(x = 4\).

The bottom function is \(x\). Its value and its slope at 4 are in the last two rows of the table.

Solution

  1. The top function \(m\) has the value 10 and the slope \(-6\). The bottom function \(x\) has the value 4 and the slope 1.

  2. Substitute the four numbers into the rule, and subtract.

    \[\begin{aligned}p'(4) &= \frac{(-6)(4) - (10)(1)}{4^2} \\ &= -\frac{34}{16} = -\frac{17}{8}\end{aligned}\]

  3. The slope is negative because the top falls and the bottom grows, and both make the ratio smaller.

Questions

The table gives \(f\), \(f'\), \(g\) and \(g'\) at four inputs. Let \(Q = f/g\). Questions 1 and 2 use the table. Question 3 needs a calculation.

  1. What is \(Q'(4)\)?

    • \(\tfrac43\)
    • \(-\tfrac43\)
    • 4
    • −2
    Answer

    \(\tfrac43\)

  2. What is \(Q'(3)\)?

    • It does not exist, because \(g(3) = 0\)
    • −16
    • \(-\tfrac32\)
    • 0
    Answer

    It does not exist, because \(g(3) = 0\)

  3. Let \(h(x) = \dfrac{p(x)}{x^2}\), with \(p(3) = 12\) and \(p'(3) = 5\). Find \(h'(3)\).

    • \(-\tfrac13\)
    • \(\tfrac59\)
    • \(\tfrac13\)
    • −3
    Answer

    \(-\tfrac13\)

Summary

You read f, f′, g and g′ from one column, test that g is not 0, and substitute the four numbers into the rule.

The rule needs only these numbers, because the slope of a ratio depends on the values and slopes of its top and bottom.

You use it when a problem gives a table or a few values and no formulas.

This is one of four skills in topic 2.9 of Which & Why: AP Calculus.