Question

If $g$ is the inverse function of $f$ and $f^{\prime }(x)=sinx$, then $g^{\prime }(x)$ is

• A. $cosec\left\{g\left(x\right)\right\}$
• B. $sin\left\{g\left(x\right)\right\}$
• C. $-\frac{1}{sin\left\{g\left(x\right)\right\}}$
• D. $cos\left\{g\left(x\right)\right\}$

Answer 1

Answer: (A)

Given $f^{-1}(x)=g(x)$
$\Rightarrow x=f\left[g\left(x\right)\right]$
Diff. both side $w.r.t\left(x\right)$
$\Rightarrow 1=f^{\prime }\left[g\left(x\right)\right].g^{\prime }\left(x\right)\Rightarrow g^{\prime }\left(x\right)=\frac{1}{f^{\prime }\left(g\left(x\right)\right)}$
Given, $f^{\prime }\left(x\right)=sinx$
$\therefore f^{\prime }\left(g\left(x\right)\right)=sin\left[g\left(x\right)\right]$
$\Rightarrow \frac{1}{f^{\prime }\left(g\left(x\right)\right)}=cosec\left[g\left(x\right)\right]$
Hence, $g^{\prime }\left(x\right)=cosec\left[g\left(x\right)\right]$