   Question Answers

# 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

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.

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 $f'\left\{ {g\left( x \right)} \right\}$
$\Rightarrow f'\left\{ {g\left( x \right)} \right\} = \sin \left\{ {g\left( x \right)} \right\}........................\left( 3 \right)$
$\Rightarrow \sin \left\{ {g\left( x \right)} \right\} \times g'\left( x \right) = 1 \\ \Rightarrow g'\left( x \right) = \dfrac{1}{{\sin \left\{ {g\left( x \right)} \right\}}} = \operatorname{cosec} \left\{ {g\left( x \right)} \right\}{\text{ }}\left[ {\because \operatorname{cosec} x = \dfrac{1}{{\sin x}}} \right] \\ \therefore g'\left( x \right) = \operatorname{cosec} \left\{ {g\left( x \right)} \right\} \\$
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 $g\left( y \right) = x$.