If $\mathrm{f}(x)=\sin ^{-1}\left(\frac{2 \log x}{1+(\log x)^2}\right)$, then…
If $\mathrm{f}(x)=\sin ^{-1}\left(\frac{2 \log x}{1+(\log x)^2}\right)$, then $\mathrm{f}^{\prime}(\mathrm{e})$ is
- $\frac{2}{\mathrm{e}}$
- $\frac{1}{2 \mathrm{e}}$
- $e$
- $\frac{1}{\mathrm{e}}$
Solution
$\begin{aligned} & f(x)=\sin ^{-1}\left(\frac{2 \log x}{1+(\log x)^2}\right) \\ \therefore \quad f^{\prime}(x) & =\frac{1}{\sqrt{1-\left(\frac{2 \log x}{1+(\log x)^2}\right)^2}}\end{aligned}$
$\begin{aligned} & =\frac{1+(\log x)^2}{\sqrt{1+(\log x)^4+2(\log x)^2-4(\log x)^2}} \\ & \times \frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{2 \log x}{1+(\log x)^2}\right) \\ & =\frac{1+(\log x)^2}{\sqrt{1-2(\log x)^2+(\log x)^4}} \\ & \times \frac{\left[1+(\log x)^2\right] \times \frac{2}{x}-(2 \log x)\left(\frac{2 \log x}{x}\right)}{\left[1+(\log x)^2\right]^2} \\ & =\frac{1}{1-(\log x)^2} \times \frac{2+2(\log x)^2-4(\log x)^2}{x\left[1+(\log x)^2\right]} \\ & =\frac{1}{1-(\log x)^2} \times \frac{2\left[1-(\log x)^2\right]}{x\left[1+(\log x)^2\right]} \\ & =\frac{2}{x\left[1+(\log x)^2\right]} \\ & \therefore \quad \mathrm{f}^{\prime}(\mathrm{e})=\frac{1}{\mathrm{e}} \\ & \end{aligned}$
Asked in: MHT CET 2023 (12 May Shift 2)
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