Question

Let$\mathrm{g}\left(\mathrm{x}\right)$ be the inverse of an invertible function $\mathrm{f}\left(\mathrm{x}\right)$ which is differentiable at$\mathrm{x}=\mathrm{c}$, then $\mathrm{g}’\left(\mathrm{f}\right(\mathrm{c}\left)\right)$ equals

• A. $\mathrm{f}’\left(\mathrm{c}\right)$
• B. $\frac{1}{\mathrm{f}’\left(\mathrm{c}\right)}$
• C. $\mathrm{f}\left(\mathrm{c}\right)$
• D. None of the above

Answer 1

The correct option is B

$\frac{1}{\mathrm{f}’\left(\mathrm{c}\right)}$

Determine the value of $\mathrm{g}’\left(\mathrm{f}\right(\mathrm{c}\left)\right)$

We are given that $\mathrm{g}\left(\mathrm{x}\right)$ is the inverse of an invertible function $\mathrm{f}\left(\mathrm{x}\right)$ which is differentiable at $\mathrm{x}=\mathrm{c}$. Therefore,

$$\begin{gathered} \mathrm{g}\left(\mathrm{x}\right)={\mathrm{f}}^{-1}\left(\mathrm{x}\right) \\ \Rightarrow \mathrm{f}\left(\mathrm{x}\right)={\mathrm{g}}^{-1}\left(\mathrm{x}\right) \\ \Rightarrow \mathrm{g}\left(\mathrm{f}\right(\mathrm{x}\left)\right)=\mathrm{x} \\ \Rightarrow gof\left(\mathrm{x}\right)=\mathrm{x} \end{gathered}$$

Now differentiate it with respect to $\mathrm{x}$ we get,

$$\begin{gathered} \Rightarrow {\mathrm{gof}}^{,}\left(\mathrm{x}\right)=1 \\ \Rightarrow {\mathrm{g}}^{,}\left(\mathrm{f}\right(\mathrm{x}\left)\right)\mathrm{f}\text{'}\left(\mathrm{x}\right)=1(\because \text{chain rule}) \\ \Rightarrow \mathrm{g}\text{'}\left(\mathrm{f}\right(\mathrm{x}\left)\right)=\frac{1}{\mathrm{f}\text{'}\left(\mathrm{x}\right)} \end{gathered}$$

Since the function is differentiable at $\mathrm{x}=\mathrm{c}$, hence replace $\mathrm{x}$ by $\mathrm{c}$ we get,

$\mathrm{g}\text{'}\left(\mathrm{f}\right(\mathrm{c}\left)\right)=\frac{1}{\mathrm{f}\text{'}\left(\mathrm{c}\right)}$

Hence, option B is the correct answer.