Differentiate inverse sine and inverse cosine

The slope of the angle whose sine is x

The green curve is \(y = \arcsin x\), and it is green because it rises. The input \(x\) is a sine value, so it lies between \(-1\) and 1. The output \(y\) is the angle between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\) whose sine is \(x\).

The orange point is at \(x = 0.6\). The triangle at the left has hypotenuse 1 and the angle \(y\), so the side opposite the angle is \(\sin y = 0.6\). The teal side next to the angle is 0.8, because \(0.6^2 + 0.8^2 = 1\).

The orange tangent has slope 1.25, which is 1 divided by 0.8. Drag the point up the curve or move the slider. The teal side becomes shorter, so the slope grows.

One over the side next to the angle

Because \(\sin y = x\), the green curve is the sine curve with \(x\) and \(y\) swapped, which is its reflection in the faint diagonal \(y = x\). The hypotenuse is 1, so the side next to the angle is \(\cos y\).

A reflection swaps the rise and the run of every tangent. This means the slope of the green curve is 1 divided by the slope of the sine curve at the matching point, which is \(\cos y\).

The angle \(y\) lies between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\), where cosine is positive, so the slope is positive.

Notation

1. The angle \(y\) is the inverse sine of \(x\) when \(\sin y = x\) and \(y\) lies between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\).

   \[y = \arcsin x \iff \sin y = x\]

2. Differentiate both sides with respect to \(x\). The angle \(y\) depends on \(x\), so by the chain rule the derivative of \(\sin y\) is \(\cb{\cos y}\) times \(\frac{dy}{dx}\).

   \[\cb{\cos y} \cdot \frac{dy}{dx} = 1\]

3. You can divide by \(\cos y\), because the side next to the angle is not 0 when \(-1 < x < 1\).

   \[\frac{dy}{dx} = \frac{1}{\cb{\cos y}}\]

4. To write \(\cos y\) in terms of \(x\), solve \(\sin^2 y + \cos^2 y = 1\) for \(\cos y\). Take the positive root, because \(\cos y > 0\) for \(-\frac{\pi}{2} < y < \frac{\pi}{2}\).

   \[\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}\]

5. The result is the derivative of the inverse sine.

   \[\begin{aligned} \frac{d}{dx} \arcsin x &= \cy{\frac{1}{\sqrt{1 - x^2}}} \\ &\text{for } -1 < x < 1 \end{aligned}\]

6. The other angle of the triangle is \(\frac{\pi}{2} - y\), and its cosine is \(x\), so \(\arccos x = \frac{\pi}{2} - \arcsin x\). The constant \(\frac{\pi}{2}\) has derivative 0, so only the sign changes.

   \[\frac{d}{dx} \arccos x = \cy{-\frac{1}{\sqrt{1 - x^2}}}\]

When it fails

As \(x\) comes close to 1, the side next to the angle becomes shorter and shorter, so the slope grows without limit. At \(x = 1\), the tangent is vertical, so the red label says there is no slope.

The formula needs \(-1 < x < 1\), because at \(x = 1\) and \(x = -1\) the function has a value but no derivative.

Example 1: inverse sine at a point

Find the slope of \(y = \arcsin x\) at \(x = 0.8\).

1. The hypotenuse is 1 and the side opposite the angle is 0.8, so the side next to the angle is the square root of \(1 - 0.8^2\).

   \[\sqrt{1 - 0.8^2} = \sqrt{0.36} = \cb{0.6}\]

2. The slope is 1 divided by the side next to the angle.

   \[\frac{d}{dx} \arcsin x \Big|_{x=0.8} = \frac{1}{0.6} = \cy{\frac{5}{3}}\]

3. As a check, \(\frac{5}{3} > 1\), and the curve in the figure is steeper at \(x = 0.8\) than at the origin, where its slope is 1.

Example 2: inverse cosine at a point

Find the slope of \(y = \arccos x\) at \(x = \dfrac{7}{25}\).

1. The curve is red because it falls from height \(\pi\) at \(x = -1\) to height 0 at \(x = 1\). At \(x = \frac{7}{25}\), the root \(\sqrt{1 - x^2}\) is \(\frac{24}{25}\).

   \[\sqrt{1 - \frac{49}{625}} = \sqrt{\frac{576}{625}} = \frac{24}{25}\]

2. A tempting wrong move: Use the same formula as for the inverse sine.

   \[\begin{aligned} \frac{d}{dx} \arccos x &= \cr{+\frac{1}{\sqrt{1 - x^2}}} \\ &= \cr{\frac{25}{24}} \end{aligned}\]

   The dashed red line rises, but the curve falls, so the slope at the point is negative and \(\frac{25}{24}\) cannot be correct.

3. The derivative of the inverse cosine has a minus sign. Divide \(-1\) by the root.

   \[\begin{aligned} \frac{d}{dx} \arccos x \Big|_{x=7/25} &= -\frac{1}{24/25} \\ &= \cy{-\frac{25}{24}} \end{aligned}\]

Practice

Find the slope of \(y = \arcsin x\) at \(x = \dfrac{5}{13}\).

Move the slider until the side opposite the angle is 5/13. Then use the hypotenuse 1 to find the side next to the angle.

Questions

The figure shows the curve each question names, with an orange point at its input. The inverse sine rises, so it is green. The inverse cosine falls, so it is red.

1. What is the slope of \(y = \arcsin x\) at \(x = 0.6\)?

   • \(\tfrac{5}{4}\)
   • \(\tfrac{4}{5}\)
   • \(\tfrac{1}{\sqrt{0.4}}\)
   • \(\tfrac{5}{3}\)
   Answer

   \(\tfrac{5}{4}\)

2. What is the slope of \(y = \arccos x\) at \(x = -0.6\)?

   • \(-\tfrac{5}{4}\)
   • \(\tfrac{5}{4}\)
   • \(-\tfrac{4}{5}\)
   Answer

   \(-\tfrac{5}{4}\)

3. What is the slope of \(y = \arcsin x\) at \(x = \tfrac{1}{2}\)?

   • \(\tfrac{2}{\sqrt{3}}\)
   • \(\sqrt{2}\)
   • \(\tfrac{4}{3}\)
   • \(-\tfrac{2}{\sqrt{3}}\)
   Answer

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

Summary

The derivative of \(\arcsin x\) is \(\frac{1}{\sqrt{1 - x^2}}\), and the derivative of \(\arccos x\) is the same with a minus sign.

This works because the slope is 1 divided by the side next to the angle, and the inverse cosine falls where the inverse sine rises.

The formulas hold for \(-1 < x < 1\), where the tangent is not vertical.

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