Evaluate: frac{d}{d x}left[e^{tan ^{-1} x+cot | vedclass

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(A) Let $y = e^{	an ^{-1} x+\cot ^{-1} x}+\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}$.We know that $	an ^{-1} x + \cot ^{-1} x

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$\sqrt{a^2-x^2}$

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(A) Let $y = e^{	an ^{-1} x+\cot ^{-1} x}+\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}$.We know that $	an ^{-1} x + \cot ^{-1} x = \frac{\pi}{2}$ for all $x \in \mathbb{R}$.So,the first term becomes $e^{\pi/2}$,which is a constant. The derivative of a constant is $0$.Now,differentiate the remaining terms: $\frac{d}{dx} \left[ \frac{x}{2} \sqrt{a^2-x^2} + \frac{a^2}{2} \sin ^{-1} \frac{x}{a} ight]$.Using the product rule for the first part: $\frac{d}{dx} \left( \frac{x}{2} \sqrt{a^2-x^2} ight) = \frac{1}{2} \sqrt{a^2-x^2} + \frac{x}{2} \cdot \frac{1}{2\sqrt{a^2-x^2}} \cdot (-2x) = \frac{1}{2} \sqrt{a^2-x^2} - \frac{x^2}{2\sqrt{a^2-x^2}} = \frac{a^2-x^2-x^2}{2\sqrt{a^2-x^2}} = \frac{a^2-2x^2}{2\sqrt{a^2-x^2}}$.For the second part: $\frac{d}{dx} \left( \frac{a^2}{2} \sin ^{-1} \frac{x}{a} ight) = \frac{a^2}{2} \cdot \frac{1}{\sqrt{1-(x/a)^2}} \cdot \frac{1}{a} = \frac{a^2}{2} \cdot \frac{1}{\sqrt{(a^2-x^2)/a^2}} \cdot \frac{1}{a} = \frac{a^2}{2} \cdot \frac{a}{\sqrt{a^2-x^2}} \cdot \frac{1}{a} = \frac{a^2}{2\sqrt{a^2-x^2}}$.Adding these results: $\frac{a^2-2x^2}{2\sqrt{a^2-x^2}} + \frac{a^2}{2\sqrt{a^2-x^2}} = \frac{2a^2-2x^2}{2\sqrt{a^2-x^2}} = \frac{2(a^2-x^2)}{2\sqrt{a^2-x^2}} = \sqrt{a^2-x^2}$.

Evaluate: $\frac{d}{d x}\left[e^{\tan ^{-1} x+\cot ^{-1} x}+\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}\right]$

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