frac{d}{dx} left( tan^{-1} frac{x}{sqrt{a^2 - | vedclass

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(D) Let $y = 	an^{-1} \left( \frac{x}{\sqrt{a^2 - x^2}} ight)$.Substitute $x = a \sin 	heta$,which implies $	heta = \sin^{-1} \left( \frac{x}{a}

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$\frac{1}{\sqrt{a^2 - x^2}}$

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(D) Let $y = 	an^{-1} \left( \frac{x}{\sqrt{a^2 - x^2}} ight)$.Substitute $x = a \sin 	heta$,which implies $	heta = \sin^{-1} \left( \frac{x}{a} ight)$.Then,$\sqrt{a^2 - x^2} = \sqrt{a^2 - a^2 \sin^2 	heta} = a \cos 	heta$.Substituting these into the expression:$y = 	an^{-1} \left( \frac{a \sin 	heta}{a \cos 	heta} ight) = 	an^{-1} (	an 	heta) = 	heta$.Thus,$y = \sin^{-1} \left( \frac{x}{a} ight)$.Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx} \left( \sin^{-1} \left( \frac{x}{a} ight) ight) = \frac{1}{\sqrt{1 - (x/a)^2}} \cdot \frac{1}{a} = \frac{1}{\sqrt{\frac{a^2 - x^2}{a^2}}} \cdot \frac{1}{a} = \frac{a}{\sqrt{a^2 - x^2}} \cdot \frac{1}{a} = \frac{1}{\sqrt{a^2 - x^2}}$.

$\frac{d}{dx} \left( \tan^{-1} \frac{x}{\sqrt{a^2 - x^2}} \right) = $

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