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)

cosec\left\\{g\left(x\right)\right\\}

Answer 2

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, $g '\left(x\right) = cosec\left[g\left(x\right)\right]$