Question

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

• A. $\csc \left\{g\right(x\left)\right\}$
• 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 A $\csc \left\{g\right(x\left)\right\}$

$f^{{}^{\prime }}\left(x\right)=\sin x$

$f\left(x\right)=-\cos x+c$ (Integrating both sides w.r.t x)

Substitute $f\left(x\right)=y$

$y=-\cos x+c$

$x={\cos }^{-1}(c-y)$

$f^{-1}y={\cos }^{-1}(c-y)$

or $g\left(x\right)={\cos }^{-1}(c-x)$

$g^{\prime }\left(x\right)=\frac{1}{\sqrt{1-(c-x{)}^2}}$

$g^{\prime }\left(x\right)=\frac{1}{\sqrt{1-{\cos }^2\left(g\right(x\left)\right)}}$

$\Rightarrow g^{\prime }\left(x\right)=\frac{1}{\sin g\left(x\right)}$

$\Rightarrow g^{\prime }\left(x\right)=\csc g\left(x\right)$