Differentiation of inverses

A clever use of the chain rule arises when we have a function \(f\) and its inverse function \(f^{-1}\). Refer to the module Functions II for a discussion of inverse functions.

Letting \(y = f(x)\), we can express \(x\) as the inverse function of \(y\):

\[ y = f(x), \qquad x = f^{-1}(y). \]

The composition of \(f\) and its inverse \(f^{-1}\), by definition, is just \(x\). That is, \(f^{-1}(f(x)) = x\). We can think of this diagrammatically as

\[ x \stackrel{f}{\to} y \stackrel{f^{-1}}{\to} x. \]

Using the chain rule, we can differentiate this composition of functions to obtain

\[ \dfrac{dx}{dx} = \dfrac{dx}{dy} \, \dfrac{dy}{dx}. \]

The derivative \(\dfrac{dx}{dx}\) of \(x\) with respect to \(x\) is just 1, so we obtain the important formula

\[ 1 = \dfrac{dx}{dy} \, \dfrac{dy}{dx}, \]

which can also be expressed as

\[ \dfrac{dy}{dx} = \dfrac{1}{\dfrac{dx}{dy}} \qquad\text{or}\qquad \dfrac{dx}{dy} = \dfrac{1}{\dfrac{dy}{dx}}. \]

This formula allows us to differentiate inverse functions, as in the following example.


Let \(y = \sqrt[3]{x}\). Find \(\dfrac{dy}{dx}\).


(We assume that we only know the derivative of \(x^n\) when \(n\) is a positive integer.) The inverse function of the cube-root function is the cube function: if \(y = \sqrt[3]{x}\), then \(x = y^3\). We know the derivative of the cube function, \(\dfrac{dx}{dy} = 3y^2\). We use this to find \(\dfrac{dy}{dx}\): \[ \dfrac{dy}{dx} = \dfrac{1}{\dfrac{dx}{dy}} = \dfrac{1}{3y^2}. \]

Substituting \(y = \sqrt[3]{x} = x^{\frac{1}{3}}\) gives

\[ \dfrac{dy}{dx} = \dfrac{1}{3 x^{\frac{2}{3}}} = \dfrac{1}{3} x^{-\frac{2}{3}}, \]

as expected.

The following exercises generalise the above example to find the derivative of \(\sqrt[n]{x} = x^{\frac{1}{n}}\), and then of any rational power of \(x\).

Exercise 20

Let \(n\) be a positive integer. Using the fact that the derivative of \(x^n\) is \(nx^{n-1}\), prove that the derivative of \(x^{\frac{1}{n}}\) is \(\dfrac{1}{n} x^{\frac{1}{n} - 1}\).

Exercise 21

Using the chain rule and the previous exercise, prove that for any rational number \(\dfrac{p}{q}\) the derivative of \(x^{\frac{p}{q}}\) is \(\dfrac{p}{q} x^{\frac{p}{q}-1}\). (Here \(p,q\) are integers and \(q > 0\).)

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