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

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

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

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

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