Q47 / 125 · BITSAT 2015 · Mathematics

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

  • A
    $\operatorname{cosec}\{\mathrm{g}(\mathrm{x})\}$
  • B
    $\sin \{g(x)\}$
  • Correct answer
    C
    $\frac{1}{\sin \{g(x)\}}$
  • D
    None of these

Solution

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

Hence, the correct answer is option: 'C'.

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