At $x=\frac{\pi^2}{4}, \frac{d}{d x}\left(\tan ^{-1}(\cos \sqrt{x})+\sec ^{-1}\left(e^x\right)\right)=$
At $x=\frac{\pi^2}{4}, \frac{d}{d x}\left(\tan ^{-1}(\cos \sqrt{x})+\sec ^{-1}\left(e^x\right)\right)=$
- $\frac{1}{\sqrt{e^{\frac{\pi^2}{2}}-1}}-\frac{1}{\pi}$
- $\frac{\pi}{4}+\frac{1}{\sqrt{\mathrm{e}^{\pi^2}+\mathrm{e}^{\pi^2 / 2}}}$
- $\frac{1}{\sqrt{\mathrm{e}^{\pi^2}+\mathrm{e}^{\pi^2 / 2}}}+\frac{2}{\pi} \cot \left(\frac{\sqrt{\pi}}{2}\right)$
- $\frac{1}{\sqrt{\mathrm{e}^\pi}}+\frac{1}{\pi}$
Solution
$
\begin{aligned}
& \text { } \frac{d}{d x}\left(\tan ^{-1}(\cos \sqrt{x})+\sec ^{-1}\left(\mathrm{e}^x\right)\right) \\
& =\frac{(-\sin \sqrt{x})}{1+\cos ^2 \sqrt{x}} \cdot\left(\frac{1}{2 \sqrt{x}}\right)+\frac{1}{e^x \sqrt{e^{2 x}-1}} \cdot e^x
\end{aligned}
$
but when $x=\frac{\pi^2}{4}$
$
\begin{aligned}
& =\frac{-\sin \frac{\pi}{2}}{1+\cos ^2 \frac{\pi}{2}}\left(\frac{1}{2}\right)\left(\frac{2}{\pi}\right)+\frac{1}{\sqrt{e^{\pi 2 / 2}-1}} \\
& =-\frac{1}{\pi}+\frac{1}{\sqrt{e^{\pi^2 / 2}-1}}
\end{aligned}
$
Asked in: AP EAMCET 2023 (19 May Shift 1)
Practice more Differentiation questions on Aicharya