Question

Derivatives of Inverse Functions Suppose that \(f\) is a one-to-one differentiable function and its inverse function \({f^{ - 1}}\) is also differentiable. Use implicit differentiation to show that

\({\left( {{f^{ - 1}}} \right)^\prime }\left( x \right) = \frac{1}{{f'\left( {{f^{ - 1}}\left( x \right)} \right)}}\)

Provided that the denominator is not 0.

Short answer

It is proved that \({\left( {{f^{ - 1}}} \right)^\prime }\left( x \right) = \frac{1}{{f'\left( {{f^{ - 1}}\left( x \right)} \right)}}\).

Step-by-step solution

 Step 1: Inverse function

It is known that if \(y = {f^{ - 1}}\left( x \right)\) is an inverse function, then \(f\left( y \right) = x\).

Implicit differentiation

Differentiate \(f\left( y \right) = x\) implicitly with respect to \(x\) as follows:

\(\begin{aligned}\frac{d}{{dx}}\left( {f\left( y \right)} \right) & = \frac{d}{{dx}}\left( x \right)\\f'\left( y \right)\frac{d}{{dx}}\left( y \right) & = 1\\\frac{d}{{dx}}\left( {{f^{ - 1}}\left( x \right)} \right) & = \frac{1}{{{f^{ - 1}}\left( y \right)}}\\{\left( {{f^{ - 1}}} \right)^\prime }\left( x \right) & = \frac{1}{{f'\left( {{f^{ - 1}}\left( x \right)} \right)}}\end{aligned}\)

Hence, proved that \({\left( {{f^{ - 1}}} \right)^\prime }\left( x \right) = \frac{1}{{f'\left( {{f^{ - 1}}\left( x \right)} \right)}}\).