## Content

### 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.

#### Example

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

#### Solution

(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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