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

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 $g\left( y \right) = x$.