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

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(D) Given $y = \sec(	an^{-1} x)$.Using the chain rule,$\frac{dy}{dx} = \sec(	an^{-1} x) 	an(	an^{-1} x) \cdot \frac{d}{dx}(	an^{-1} x)$.Since $\f

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$\frac{1}{\sqrt{2}}$

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(D) Given $y = \sec(	an^{-1} x)$.Using the chain rule,$\frac{dy}{dx} = \sec(	an^{-1} x) 	an(	an^{-1} x) \cdot \frac{d}{dx}(	an^{-1} x)$.Since $\frac{d}{dx}(	an^{-1} x) = \frac{1}{1+x^2}$,we have $\frac{dy}{dx} = \sec(	an^{-1} x) \cdot x \cdot \frac{1}{1+x^2}$.At $x = 1$,$	an^{-1}(1) = \frac{\pi}{4}$.So,$\frac{dy}{dx} = \sec(\frac{\pi}{4}) \cdot 1 \cdot \frac{1}{1+1^2} = \sqrt{2} \cdot \frac{1}{2} = \frac{1}{\sqrt{2}}$.Alternatively,let $	an^{-1} x = 	heta$,then $	an 	heta = x$. Since $\sec^2 	heta = 1 + 	an^2 	heta$,we have $\sec 	heta = \sqrt{1+x^2}$.Thus,$y = \sqrt{1+x^2}$.Then $\frac{dy}{dx} = \frac{1}{2\sqrt{1+x^2}} \cdot 2x = \frac{x}{\sqrt{1+x^2}}$.At $x = 1$,$\frac{dy}{dx} = \frac{1}{\sqrt{1+1^2}} = \frac{1}{\sqrt{2}}$.

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

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