Question

If $g$ is the inverse function of $f$ and $f '(x) = sin\, x$, then $g '(x)$ 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 $w.r.t \left(x\right)$
$\Rightarrow 1=f '\left[g\left(x\right)\right].g '\left(x\right) \Rightarrow g '\left(x\right)=\frac{1}{f '\left(g\left(x\right)\right)}$
Given, $f '\left(x\right) = sin\, x$
$\therefore f '\left(g\left(x\right)\right)=sin\left[g\left(x\right)\right]$
$\Rightarrow \frac{1}{f '\left(g\left(x\right)\right)}=co\,sec\left[g\left(x\right)\right]$
Hence, $g '\left(x\right) = cosec\left[g\left(x\right)\right]$