Question

If \[g\] is the inverse of the function \[f\] and \[f'\left( x \right) = \sin x\], then \[g'\left( x \right) = \]
A. \[\operatorname{cosec} \left\{ {g\left( x \right)} \right\}\]
B. \[\sin \left\{ {g\left( x \right)} \right\}\]
C. \[ - \dfrac{1}{{\sin \left\{ {g\left( x \right)} \right\}}}\]
D. None of these

Answer 1

Hint: In this question, we will proceed by finding a relation between \[g\] and \[f\]. Then find the derivative of the obtained equation w.r.t \[x\]. Further use the conversion \[\operatorname{cosec} x = \dfrac{1}{{\sin x}}\] to get the required answer. So, use this concept to reach the solution of the given problem.

Complete step-by-step answer:
Given that \[g\] is the inverse of the function \[f\] i.e., \[g\left( x \right) = {f^{ - 1}}\left( x \right)\].
So, we have \[f\left\{ {g\left( x \right)} \right\} = x..........................................\left( 1 \right)\]
Differentiating equation (1) w.r.t ‘\[x\]’, we have
\[
   \Rightarrow \dfrac{d}{{dx}}\left[ {f\left\{ {g\left( x \right)} \right\}} \right] = \dfrac{d}{{dx}}\left( x \right) \\
   \Rightarrow f'\left\{ {g\left( x \right)} \right\} \times g'\left( x \right) = 1..............................\left( 2 \right) \\
\]
But give that \[f'\left( x \right) = \sin x\]
Now, consider \[f'\left\{ {g\left( x \right)} \right\}\]
\[ \Rightarrow f'\left\{ {g\left( x \right)} \right\} = \sin \left\{ {g\left( x \right)} \right\}........................\left( 3 \right)\]
From equation (2) and (3), we have
\[
   \Rightarrow \sin \left\{ {g\left( x \right)} \right\} \times g'\left( x \right) = 1 \\
   \Rightarrow g'\left( x \right) = \dfrac{1}{{\sin \left\{ {g\left( x \right)} \right\}}} = \operatorname{cosec} \left\{ {g\left( x \right)} \right\}{\text{ }}\left[ {\because \operatorname{cosec} x = \dfrac{1}{{\sin x}}} \right] \\
  \therefore g'\left( x \right) = \operatorname{cosec} \left\{ {g\left( x \right)} \right\} \\
\]
Thus, the correct option is A. \[\operatorname{cosec} \left\{ {g\left( x \right)} \right\}\]

Note: In mathematics, an inverse function (or anti-function0 is a function that reverts another function. For example if a function \[f\] applied to an input \[x\] gives a result of \[y\], then applying its inverse function \[g\] to \[y\] gives the result \[x\], and vice-versa, i.e., \[f\left( x \right) = y\] if and only if \[g\left( y \right) = x\].