Check where the derivative is defined
A curve that stops at two vertical tangents
The curve is \(y = \arcsin(2x - 1)\), and it is green because it rises. The inverse sine accepts only values from \(-1\) to 1. The inner value \(u = 2x - 1\) is in that range only for \(x\) from 0 to 1, so the curve exists only on that interval.
At the orange point, \(x = \frac{1}{2}\) and \(u = 0\), and the tangent has slope 2.
Drag the point toward \(x = 1\). The tangent becomes steeper, and at the end point it is vertical, so the slope does not exist there.
The function exists from 0 to 1, end points included, but its slope exists only between them.
Two domains
The curve exists where the inner value \(u\) is from \(-1\) to 1. The solid dots mark its end points, where \(u = -1\) and \(u = 1\).
The slope exists only where \(u\) is strictly between \(-1\) and 1. The line under the curve, marked \(f'\), shows this interval, and its open rings leave out the end points. It is green because the slope is positive.
The two intervals differ only by the end points, where the tangent is vertical.
Notation
Let \(f(x) = \arcsin(u(x))\). The inverse sine accepts values from \(-1\) to 1, ends included, so \(f\) exists where this holds.
\[-1 \le u(x) \le 1\]Differentiate with the chain rule.
\[f'(x) = \frac{u'(x)}{\sqrt{1 - u(x)^2}}\]The square root in the denominator cannot be 0, so \(1 - u^2\) must be positive. This excludes both ends.
\[-1 < u(x) < 1\]The inner derivative \(u'\) is in the numerator, so it must also exist. Remove the points where it does not.
The inverse tangent has no such restriction, because it accepts every value of \(u\) and \(1 + u^2\) is never 0. Only the points where \(u'\) does not exist are removed.
\[\frac{d}{dx} \arctan u = \frac{u'}{1 + u^2}\]
When it fails
At \(x = 1\) the inner value is \(u = 1\), and the function has the value \(\frac{\pi}{2}\). The formula divides by \(\sqrt{1 - 1} = 0\).
The tangent is vertical, so the derivative does not exist. A function can exist at a point where its derivative does not.
Example 1: inverse cosine of x over 3
Find the domain of \(f(x) = \arccos\left(\dfrac{x}{3}\right)\) and the domain of \(f'\).
The curve is red because it falls. The inverse cosine accepts values from \(-1\) to 1, so multiply each part by 3 to find \(x\).
\[-1 \le \frac{x}{3} \le 1 \ \Rightarrow\ \cy{-3 \le x \le 3}\]Differentiate. The inner derivative is \(\frac{1}{3}\), and the derivative of the inverse cosine has a minus sign. To simplify, multiply the top and the bottom by \(3 = \sqrt{9}\).
\[\begin{aligned} f'(x) &= -\frac{1/3}{\sqrt{1 - x^2/9}} \\ &= -\frac{1}{\sqrt{9 - x^2}} \end{aligned}\]The square root in the denominator must be positive, so \(9 - x^2 > 0\), which excludes \(x = -3\) and \(x = 3\). The line under the curve shows this open interval, and it is red because \(f'\) is negative there.
\[\cy{-3 < x < 3}\]At both ends the tangent is vertical, so \(f'\) exists only on the open interval.
Example 2: inverse sine of a square root
Find the domain of \(f(x) = \arcsin\left(\sqrt{x}\right)\) and the domain of \(f'\).
The square root needs \(x \ge 0\). The inverse sine needs \(\sqrt{x} \le 1\), which means \(x \le 1\).
\[\cy{0 \le x \le 1}\]Differentiate with \(u = \sqrt{x}\) and \(u' = \frac{1}{2\sqrt{x}}\). Because \(u^2 = x\), the root in the denominator is \(\sqrt{1 - x}\).
\[\begin{aligned} f'(x) &= \frac{1/(2\sqrt{x})}{\sqrt{1 - x}} \\ &= \frac{1}{2\sqrt{x(1 - x)}} \end{aligned}\]A tempting wrong move: The function exists on the closed interval, so the derivative exists there too.
\[\text{domain of } f' = \cr{[0, 1]}\]At \(x = 0.99\), the slope is already about 5, and the tangent at \(x = 1\) is vertical, so the slope does not exist there.
The denominator is 0 at both ends. At \(x = 1\), the inner value is 1, and at \(x = 0\), the inner derivative \(\frac{1}{2\sqrt{x}}\) does not exist.
\[\cy{0 < x < 1}\]
Practice
Find the domain of \(f(x) = \arcsin(2x^2 - 1)\) and the domain of \(f'\).
Find where 2x² − 1 equals −1 or 1, because the denominator is 0 there. Then drag the point to those values of x.
Solution
The inverse sine needs \(-1 \le 2x^2 - 1 \le 1\). Add 1 to each part. Then divide each part by 2.
\[0 \le x^2 \le 1 \ \Rightarrow\ \cy{-1 \le x \le 1}\]Differentiate, with \(u = 2x^2 - 1\) and \(u' = 4x\). The expression under the root in the denominator simplifies.
\[\begin{aligned} 1 - u^2 &= 4x^2 - 4x^4 \\ &= 4x^2(1 - x^2) \end{aligned}\]The denominator is 0 at \(x = 0\), where \(u = -1\), and at \(x = \pm 1\), where \(u = 1\). The graph has a corner at \(x = 0\), where it changes from falling (red) to rising (green), so there is no tangent. Remove these three points.
\[\cy{(-1, 0) \cup (0, 1)}\]
Questions
The figure shows \(f(x) = \arcsin(3x)\), which exists on the interval from \(-\frac{1}{3}\) to \(\frac{1}{3}\). Solid dots mark its two end points.
What is true at \(x = \tfrac{1}{3}\)?
- \(f\) exists, and \(f'\) does not exist
- \(f'(\tfrac{1}{3}) = 0\)
- \(f'(\tfrac{1}{3})\) exists, because \(f\) exists
- Neither \(f\) nor \(f'\) exists
Answer
\(f\) exists, and \(f'\) does not exist
What is the domain of \(f'\)?
- \(\left(-\tfrac{1}{3}, \tfrac{1}{3}\right)\)
- \(\left[-\tfrac{1}{3}, \tfrac{1}{3}\right]\)
- \((-1, 1)\)
- All real numbers
Answer
\(\left(-\tfrac{1}{3}, \tfrac{1}{3}\right)\)
What is the domain of \(g'\) for \(g(x) = \arctan\left(\sqrt{x}\right)\)?
- \((0, \infty)\)
- \([0, \infty)\)
- All real numbers
- \((0, 1)\)
Answer
\((0, \infty)\)
Summary
An inverse sine or inverse cosine \(f\) exists where \(-1 \le u \le 1\), but \(f'\) exists only where \(-1 < u < 1\) and \(u'\) exists.
This works because the inverse sine and the inverse cosine have vertical tangents where \(u = \pm 1\), and there the formula divides by 0.
Use it before you state where a derivative formula is valid.
This is one of four skills in topic 3.4 of Which & Why: AP Calculus.