Question

If $g$ is the inverse of function of $f$ and $f^{\prime }\left(x\right)=\sin x$, then $g^{\prime }\left(x\right)$ is equal to

• A. $\cos \left\{g\right(x\left)\right\}$
• B. $\sin \left\{g\right(x\left)\right\}$
• C. $\frac{1}{\sin \left\{g\right(x\left)\right\}}$
• D. None of these

Answer 1

The correct option is C $\frac{1}{\sin \left\{g\right(x\left)\right\}}$

Since, $g$ is the inverse of function $f$.

Therefore, $g\left(x\right)=f^{-1}\left(x\right)\Rightarrow f\left[g\right(x\left)\right]=x$
$\Rightarrow f\circ g\left(x\right)=x$
Differentiating both side, w.r.t $x$
$\Rightarrow \frac{d}{dx}\{f\circ g(x\left)\right\}=\frac{d}{dx}\left(x\right)$ for all $x$
$\Rightarrow f^{\prime }\left[g\right(x\left)\right]g^{\prime }\left(x\right)=1$ for all $x$
$\Rightarrow \sin \left\{g\right(x\left)\right\}{g}^{\prime }\left(x\right)=1$
By definition of $f^{\prime }\left(x\right)$
$\Rightarrow g^{\prime }\left(x\right)=\frac{1}{\sin \left\{g\right(x\left)\right\}}$