Wemazi

Topic 3.3

Calculate an inverse derivative

Builds on finding the matching inverse input. 7 minutes.

Slopes at mirror points

The curve is \(f(x) = x^3/16\), and the orange point \((2, 0.5)\) lies on it. The orange line that touches the curve at this point is the tangent. The small triangle under it has rise \(\frac34\) and run 1, so its slope is \(\frac34\).

Press the button to reflect the curve, the point and the tangent across the faint diagonal line \(y = x\). The reflection swaps the coordinates of every point, so the reflected curve is the graph of the inverse and the mirror point is \((0.5, 2)\).

The reflection also swaps rise and run, so the mirror tangent has rise 1 and run \(\frac34\), and its slope is \(\frac43\).

Drag the point. Both slopes change, but the mirror slope remains the reciprocal of the slope of \(f\), so the product of the two is 1.

Reflecting swaps rise and run

At the point \((a, b)\), the tangent of \(f\) has rise \(m\) and run 1, so its slope is \(m = f'(a)\).

The reflection swaps rise and run, so the mirror tangent at \((b, a)\) has rise 1 and run \(m\), and its slope is \(\frac1m\).

Drag the point. The slope of the inverse at \(b\) remains the reciprocal of the slope of \(f\) at the matching input \(a\), not at \(b\).

Notation

  1. The function \(f\) undoes its inverse, so applying \(f\) to \(f^{-1}(x)\) returns \(x\).

    \[f\big(f^{-1}(x)\big) = x\]

  2. Differentiate both sides. The left side is a composition, so the chain rule gives the outer derivative at the inner function, times the inner derivative.

    \[f'\big(f^{-1}(x)\big) \cdot (f^{-1})'(x) = 1\]

  3. Divide both sides by \(f'(f^{-1}(x))\). You cannot divide by 0, so this step assumes that this factor is not 0.

    \[(f^{-1})'(x) = \frac{1}{f'\big(f^{-1}(x)\big)}\]

  4. At a given \(b\), write \(a\) for the matching input, where \(f(a) = b\). The slope of the inverse at \(b\) is the reciprocal of the slope of \(f\) at \(a\).

    \[\cy{(f^{-1})'(b)} = \frac{1}{\cy{f'(a)}}\]

  5. To use the rule, find the input \(a\) with \(f(a) = b\). Then take the reciprocal of \(f'(a)\).

  6. The rule needs \(f\) to be differentiable at \(a\), to have an inverse near \(a\), and to have \(f'(a) \neq 0\).

When it fails

At \(a = 0\) the tangent of \(f\) is horizontal, so \(f'(0) = 0\). The mirror tangent is vertical, and the red label marks that a vertical line has no slope.

The formula would need \(\frac10\), which does not exist, so the inverse has no derivative at \(b = 0\).

Example 1: Find the slope of an inverse

Let \(f(x) = 2x^3 + x\). Find \((f^{-1})'(3)\).

  1. Find the matching input \(a\) with \(f(a) = 3\). Try \(a = 1\).

    \[f(1) = 2 + 1 = \cy{3}\]

  2. Differentiate \(f\). Then substitute the matching input 1.

    \[f'(x) = 6x^2 + 1, \qquad f'(1) = \cy{7}\]

  3. Take the reciprocal of the slope of \(f\) at 1.

    \[(f^{-1})'(3) = \frac{1}{f'(1)} = \cy{\frac17}\]

  4. The tangent at \((1, 3)\) is steep, with slope 7, so the mirror tangent at \((3, 1)\) is nearly flat, with slope \(\frac17\).

Example 2: Read the slope at the matching input

Let \(f(x) = x^5 + x^3\). Find \((f^{-1})'(2)\).

  1. Differentiate \(f\). Both terms are even powers of \(x\) with positive coefficients, so \(f'\) is never negative, and \(f\) only rises and has an inverse.

    \[f'(x) = 5x^4 + 3x^2\]

  2. A tempting wrong move: The inverse is taken at 2, so use the slope of \(f\) at \(x = 2\).

    \[(f^{-1})'(2) = \cr{\frac{1}{f'(2)} = \frac{1}{92}}\]

    The slope \(f'(2) = 92\) belongs to the point \((2, 40)\) of \(f\), which is far above the picture, as the red line shows. The point with output 2 is \((1, 2)\), where the slope is much smaller.

  3. The inverse at 2 needs the input \(a\) with \(f(a) = 2\). Try \(a = 1\).

    \[f(1) = 1 + 1 = \cy{2}\]

  4. Substitute 1 into \(f'\). Then take the reciprocal.

    \[f'(1) = 8, \qquad (f^{-1})'(2) = \cy{\frac18}\]

Practice

Let \(f(x) = 3x + x^5\). Find \((f^{-1})'(38)\).

Find the matching input first. Move the point to where the curve reaches height 38. Then read the slope of f at that input.

Solution

  1. Find the input \(a\) with \(f(a) = 38\). Try \(a = 2\).

    \[f(2) = 6 + 32 = \cy{38}\]

  2. Differentiate \(f\). Then substitute 2. Because \(f'\) is positive for every \(x\), \(f\) only rises and has an inverse.

    \[\begin{aligned} f'(x) &= 3 + 5x^4 \\ f'(2) &= 3 + 80 = 83 \end{aligned}\]

  3. Take the reciprocal.

    \[(f^{-1})'(38) = \frac{1}{f'(2)} = \cy{\frac{1}{83}}\]

Questions

The graph shows \(f(x) = \frac{x^3}{2}\) and its mirror curve. The tangent of \(f\) at \((2, 4)\) has slope 6, and the tangent at \((-2, -4)\) also has slope 6.

  1. What is \((f^{-1})'(4)\)?

    • \(\tfrac16\)
    • 6
    • \(\tfrac{1}{24}\)
    Answer

    \(\tfrac16\)

  2. What is \((f^{-1})'(-4)\)?

    • \(\tfrac16\)
    • \(-\tfrac16\)
    • \(\tfrac{1}{24}\)
    • 6
    Answer

    \(\tfrac16\)

  3. Let \(g(x) = x + e^x\), a function that only rises. What is \((g^{-1})'(1)\)?

    • \(\tfrac12\)
    • \(\dfrac{1}{1 + e}\)
    • 2
    • 1
    Answer

    \(\tfrac12\)

Summary

To find the slope of the inverse at \(b\), you find the matching input \(a\) with \(f(a) = b\) and take the reciprocal of \(f'(a)\).

This works because a reflection across the diagonal swaps rise and run, and the chain rule gives the same result.

Use it when you need the slope of an inverse but have no formula for the inverse.

This is one of four skills in topic 3.3 of Which & Why: AP Calculus.