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If $g$ is the inverse of function $f$ and $f^{\prime}(x)=\sin x$, then $\mathrm{g}^{\prime}(\mathrm{x})$ is equal to
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Verified Answer
The correct answer is:
$\frac{1}{\sin \{g(x)\}}$
Since, $g$ is the inverse of function $f$. Therefore, $\mathrm{g}(\mathrm{x})=\mathrm{f}^{-1}(\mathrm{x})$
$\Rightarrow \mathrm{f}[\mathrm{g}(\mathrm{x})]=\mathrm{x}$
$\Rightarrow$ fog $(\mathrm{x})=\mathrm{x},$ for all $\mathrm{x}$
Differentiate both side,w.r.tx
$\Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\{\operatorname{fog}(\mathrm{x})\}=\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}),$ for all $\mathrm{x}$
$\Rightarrow \mathrm{f}^{\prime}[\mathrm{g}(\mathrm{x})] \mathrm{g}^{\prime}(\mathrm{x})=1,$ for all $\mathrm{x}$
$\Rightarrow \sin \{\mathrm{g}(\mathrm{x})\} \mathrm{g}^{\prime}(\mathrm{x})=1,$ for all $\mathrm{x}$
(By defn of $\left.f^{\prime}(x)\right)$
$\Rightarrow \mathrm{g}^{\prime}(\mathrm{x})=\frac{1}{\sin \{\mathrm{g}(\mathrm{x})\}}$
$\Rightarrow \mathrm{f}[\mathrm{g}(\mathrm{x})]=\mathrm{x}$
$\Rightarrow$ fog $(\mathrm{x})=\mathrm{x},$ for all $\mathrm{x}$
Differentiate both side,w.r.tx
$\Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\{\operatorname{fog}(\mathrm{x})\}=\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}),$ for all $\mathrm{x}$
$\Rightarrow \mathrm{f}^{\prime}[\mathrm{g}(\mathrm{x})] \mathrm{g}^{\prime}(\mathrm{x})=1,$ for all $\mathrm{x}$
$\Rightarrow \sin \{\mathrm{g}(\mathrm{x})\} \mathrm{g}^{\prime}(\mathrm{x})=1,$ for all $\mathrm{x}$
(By defn of $\left.f^{\prime}(x)\right)$
$\Rightarrow \mathrm{g}^{\prime}(\mathrm{x})=\frac{1}{\sin \{\mathrm{g}(\mathrm{x})\}}$
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