If y = sec(tan^{-1}x),then frac{dy}{dx} is | vedclass

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Question

(A) Given $y = \sec(	an^{-1}x)$.Let $	an^{-1}x = 	heta$,then $x = 	an	heta$. This implies $\sec	heta = \sqrt{1 + 	an^2	heta} = \sqrt{1 + x^2}$

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$\frac{x}{\sqrt{1 + x^2}}$

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(A) Given $y = \sec(	an^{-1}x)$.Let $	an^{-1}x = 	heta$,then $x = 	an	heta$. This implies $\sec	heta = \sqrt{1 + 	an^2	heta} = \sqrt{1 + x^2}$.So,$y = \sqrt{1 + x^2}$.Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(\sqrt{1 + x^2})$$\frac{dy}{dx} = \frac{1}{2\sqrt{1 + x^2}} \cdot \frac{d}{dx}(1 + x^2)$$\frac{dy}{dx} = \frac{1}{2\sqrt{1 + x^2}} \cdot (2x)$$\frac{dy}{dx} = \frac{x}{\sqrt{1 + x^2}}$.

If $y = \sec(\tan^{-1}x)$,then $\frac{dy}{dx}$ is

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