Wemazi

Topic 3.4

Apply the chain rule inside an inverse trig function

Builds on differentiating inverse tangent. 7 minutes.

An inverse tangent of 3x

The left graph shows the inner value \(u = 3x\), the middle graph shows \(y = \arctan u\), and the right graph shows \(y = \arctan 3x\) against \(x\). Each graph has an orange tangent with its slope.

At \(x = \frac{1}{3}\), the inner value is \(u = 1\). The slope of the middle graph, called the outer slope, is \(\frac{1}{1 + u^2} = \frac{1}{2}\).

The slope of the left graph, called the inner slope, is 3, because \(u\) changes by 3 for each unit of \(x\). Near the point, \(y\) changes half as fast as \(u\), so the right graph has slope \(3 \cdot \frac{1}{2} = \frac{3}{2}\).

Drag the point. The readouts change, but the slope of the right graph is always the outer slope times the inner slope.

The outer slope at u, times the inner slope

The inner function \(u = g(x)\) turns \(x\) into a value \(u\), and the inverse trig function \(f\) turns \(u\) into \(y\).

The outer slope \(\cy{f'(u)}\) is read on the middle graph at \(u\), and the inner slope \(\cy{g'(x)}\) is read on the left graph at \(x\).

Drag the point. Both slopes change, but the slope of the right graph is always their product \(\cy{f'(u)\,g'(x)}\).

Notation

  1. Call the input of the inverse trig function \(u\). The figure shows the inverse sine of \(u = 2x\).

    \[y = f(u), \qquad u = g(x)\]

  2. The chain rule multiplies the outer derivative, taken with respect to \(u\), by the inner derivative.

    \[\frac{dy}{dx} = \frac{dy}{du} \cdot \cy{u'}\]

  3. In the formula for the inverse tangent, put \(u\) where \(x\) was, and multiply by \(u'\).

    \[\frac{d}{dx} \arctan u = \cy{\frac{u'}{1 + u^2}}\]

  4. The inverse sine and the inverse cosine follow the same pattern, and the inverse cosine has a minus sign.

    \[\begin{aligned} \frac{d}{dx} \arcsin u &= \frac{u'}{\sqrt{1 - u^2}} \\ \frac{d}{dx} \arccos u &= -\frac{u'}{\sqrt{1 - u^2}} \end{aligned}\]

  5. To find the slope at a number, find \(u\) and \(u'\) at that number first. Then substitute both into the formula.

  6. The function \(u\) must be differentiable. For the inverse sine and the inverse cosine, \(u\) must also satisfy \(-1 < u < 1\) at the number.

When it fails

On \(y = \arcsin 2x\), the inner value \(u = 2x\) reaches 1 at \(x = \frac{1}{2}\), so \(\sqrt{1 - u^2}\) is 0 and the formula would divide by 0.

The inverse sine has no derivative at \(u = 1\), so \(\arcsin 2x\) has none at \(x = \frac{1}{2}\). Both of their tangents are vertical.

Example 1: inverse sine of x over 2

Find the slope of \(y = \arcsin\left(\dfrac{x}{2}\right)\) at \(x = 1\).

  1. Name the inner function and find its derivative.

    \[u = \frac{x}{2}, \qquad u' = \cy{\frac{1}{2}}\]

  2. At \(x = 1\), the inner value \(u = \frac{1}{2}\) lies between \(-1\) and 1, so the formula applies. Find the square root.

    \[\sqrt{1 - u^2} = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}\]

  3. Divide the inner derivative by the square root.

    \[\frac{u'}{\sqrt{1 - u^2}} = \frac{1/2}{\sqrt{3}/2} = \cy{\frac{1}{\sqrt{3}}}\]

Example 2: inverse tangent of 1 over x

Find the slope of \(y = \arctan\left(\dfrac{1}{x}\right)\) at \(x = 2\).

  1. Name the inner function and find its derivative. At \(x = 2\), the inner value is \(u = \frac{1}{2}\).

    \[u = \frac{1}{x}, \qquad u' = -\frac{1}{x^2} = \cy{-\frac{1}{4}}\]

  2. A tempting wrong move: Use the formula for the inverse tangent at \(u\), and stop.

    \[\frac{1}{1 + u^2} = \cr{\frac{1}{1 + 1/4} = \frac{4}{5}}\]

    The red line with slope \(\frac{4}{5}\) rises, but the right graph falls at \(x = 2\). The step left out the factor \(u'\), so \(\frac{4}{5}\) is only the outer slope.

  3. Multiply by the inner derivative \(u' = -\frac{1}{4}\). This factor is negative, so the slope is negative.

    \[\frac{u'}{1 + u^2} = \frac{-1/4}{5/4} = \cy{-\frac{1}{5}}\]

Practice

Find the slope of \(y = \arcsin(1 - 2x)\) at \(x = \dfrac{1}{4}\).

Move x to 1/4 and read u on the left graph. Multiply the slope of the middle graph at that u by the slope of the left graph.

Solution

  1. Name the inner function. Its value at \(x = \frac{1}{4}\) lies between \(-1\) and 1, so the formula applies.

    \[u = 1 - 2x = \frac{1}{2}, \qquad u' = \cy{-2}\]

  2. Find the square root.

    \[\sqrt{1 - u^2} = \sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}\]

  3. Divide the inner derivative by the square root. The slope is negative because \(u\) decreases as \(x\) grows.

    \[\frac{-2}{\sqrt{3}/2} = \cy{-\frac{4}{\sqrt{3}}}\]

Questions

The figure shows \(y = \arctan(x^3)\). Each question moves the orange point to the input it names.

  1. What is the slope of the tangent on the right graph at \(x = 0\)?

    • 0
    • 1
    • 3
    • The derivative does not exist
    Answer

    0

  2. What is the slope of the tangent on the right graph at \(x = 1\)?

    • \(\tfrac{3}{2}\)
    • \(\tfrac{1}{2}\)
    • 3
    • 6
    Answer

    \(\tfrac{3}{2}\)

  3. What is the slope of \(y = \arcsin(3x)\) at \(x = \tfrac{1}{6}\)?

    • \(2\sqrt{3}\)
    • \(\tfrac{2}{\sqrt{3}}\)
    • \(3\sqrt{2}\)
    • 6
    Answer

    \(2\sqrt{3}\)

Summary

You differentiate an inverse trig function of \(u\) by writing its formula at \(u\) and multiplying by \(u'\).

This works because the chain rule multiplies the outer slope, read at u, by the inner slope, read at x.

Use it when the input of an inverse trig function is a function of \(x\). Find \(u\) before you substitute the number.

This is one of four skills in topic 3.4 of Which & Why: AP Calculus.