Proposition 5.1.
For any function \(f\) so that both \(f\) and \(f^{-1}\) are differentiable, we have:
\begin{equation}
\frac{d}{dx} f^{-1} (x) = \frac{1}{f' (f^{-1} (x))} \, .\tag{5.1}
\end{equation}
Unit 5.1 Differentiating inverse functions
This section pays back a debt by addressing those functions in Proposition 4.8 whose derivations we have not yet discussed: powers, exponentials, logarithms and inverse trig functions. To clarify our terminology, the reason \(x^a\) is called a power, while \(a^x\) is called an exponential, is that we are differentiating with respect to \(x\text{,}\) while \(a\) plays the role of a constant.
For positive integer powers \(x^n\) there are many ways of computing the derivative. One is by expanding it out:
\begin{equation*}
(x+h)^n - x^n = n h x^{n-1} + {n \choose 2} h^2 x^{n-2} + \cdots +
n h^{n-1} x + h^n \, .
\end{equation*}
Dividing by \(h\) and taking the limit as \(h \to 0\) shows that the derivative of \(x^n\) is \(n x^{n-1}\text{.}\) Another way is to prove it by induction, using the product rule to get from \((d/dx) x^n = n x^{n-1}\) to \((d/dx) x^{n+1} = (n+1) x^n\text{.}\)
For negative integer powers you can use the quotient rule, writing \(x^{-n} = 1 / x^n\) and using the known derivative for positive integer values of \(n\text{.}\) For rational powers, it is easiest after proving a combining rule that tells us how to compute the derivative of the inverse function \(f^{-1}\) if we know the derivative of \(f\text{.}\) The derivation is a quick use of the chain rule.
Proof.
By definition \(f(f^{-1} (x)) = x\text{.}\) Taking the derivative of both sides,
\begin{equation*}
f' (f^{-1} (x)) \frac{d}{dx} f^{-1} (x) = 1
\end{equation*}
and dividing both sides by \(f' (f^{-1} (x))\) yields the result.
Checkpoint 90.
Suppose \(f\) has input units of people and output units of money. Do a unit analysis of (5.1): what are the units of each side, and are they the same?
Checkpoint 91.
Square root is the inverse function to squaring. Using Proposition 5.1 quickly computes the derivitive of the square root. Letting \(f(x) = x^2\) in Proposition 5.1, and using \(f'(x) = 2x\text{,}\) the conclusion becomes
\begin{equation*}
\frac{d}{dx} \sqrt{x} = \frac{1}{2 \cdot \sqrt{x}} \, .
\end{equation*}
Checkpoint 92.
Use a similar method to compute \(\displaystyle \frac{d}{dx} \sqrt[3]{x}\text{.}\) Show your work.
A very similar argument allows us to show \((d/dx) x^{1/n} = (1/n) x^{1/n - 1}\text{.}\) Using the chain rule, because \(x^{k/n} = (x^{1/n})^k\text{,}\) we can then compute \((d/dx) x^{k/n}\) for any nonzero integers \(k\) and \(n\text{.}\) So now we have verified that the derivative of \(x^r\) is \(r x^{r-1}\) for all rational numbers \(r\text{.}\)
At the end of the section we will finish this argument by handling the case of exponents that are not rational numbers.
Inverse trig functions.
We’ve already computed the derivatives of the basic trig functions (parts 6, 7 and 8 of Proposition 4.8). What remains are the inverse trig functions. Use the inverse function rule, obviously! For example, if \(f(x) := \sin x\) then the derivative of \(\arcsin\) is computed by
\begin{equation*}
\frac{d}{dx} \arcsin x = \frac{1}{\cos(\arcsin x)} \, .
\end{equation*}
Some of you may recognize the identity \(\cos (\arcsin y) = \sqrt{1 - y^2}\text{.}\) In case not, it’s an easy piece of geometry. For any \(y \in [-1,1]\text{,}\) \(\arcsin y\) is a value between \(- \pi / 2\) and \(\pi/2\text{,}\) denoted by \(\theta\) in Figure \ref{fig:arcsin}. In the figure, the measure of BC is \(|y|\) and the measure of AC is \(\cos \theta = \cos(\arcsin y)\text{,}\) and the Pythagorean theorem shows what we want, namely \(\cos \arcsin y =
\sqrt{1-y^2}\text{.}\)
