At x=frac{pi^2}{4},find the value of frac{d}{ | vedclass

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(A) Let $y = 	an^{-1}(\cos \sqrt{x}) + \sec^{-1}(e^x)$.Applying the chain rule,we differentiate with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(	a

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$\frac{1}{\sqrt{e^{\frac{\pi^2}{2}}-1}}-\frac{1}{\pi}$

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(A) Let $y = 	an^{-1}(\cos \sqrt{x}) + \sec^{-1}(e^x)$.Applying the chain rule,we differentiate with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(	an^{-1}(\cos \sqrt{x})) + \frac{d}{dx}(\sec^{-1}(e^x))$$= \frac{1}{1 + (\cos \sqrt{x})^2} \cdot (-\sin \sqrt{x}) \cdot \frac{1}{2\sqrt{x}} + \frac{1}{|e^x| \sqrt{(e^x)^2 - 1}} \cdot e^x$$= \frac{-\sin \sqrt{x}}{2\sqrt{x}(1 + \cos^2 \sqrt{x})} + \frac{1}{\sqrt{e^{2x} - 1}}$.Now,substitute $x = \frac{\pi^2}{4}$,so $\sqrt{x} = \frac{\pi}{2}$:$\frac{dy}{dx} = \frac{-\sin(\frac{\pi}{2})}{2(\frac{\pi}{2})(1 + \cos^2(\frac{\pi}{2}))} + \frac{1}{\sqrt{e^{2(\frac{\pi^2}{4})} - 1}}$$= \frac{-1}{\pi(1 + 0)} + \frac{1}{\sqrt{e^{\frac{\pi^2}{2}} - 1}}$$= \frac{1}{\sqrt{e^{\frac{\pi^2}{2}} - 1}} - \frac{1}{\pi}$.

At $x=\frac{\pi^2}{4}$,find the value of $\frac{d}{d x}\left(\tan ^{-1}(\cos \sqrt{x})+\sec ^{-1}\left(e^x\right)\right)$.

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