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

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(B) Given $y = \sec(	an^{-1} x)$.Let $	an^{-1} x = 	heta$,then $	an 	heta = x$.We know that $\sec^2 	heta = 1 + 	an^2 	heta = 1 + x^2$,so $\se

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

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(B) Given $y = \sec(	an^{-1} x)$.Let $	an^{-1} x = 	heta$,then $	an 	heta = x$.We know that $\sec^2 	heta = 1 + 	an^2 	heta = 1 + x^2$,so $\sec 	heta = \sqrt{1 + x^2}$.Thus,$y = \sqrt{1 + x^2}$.Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(\sqrt{1 + x^2}) = \frac{1}{2\sqrt{1 + x^2}} \cdot (2x) = \frac{x}{\sqrt{1 + x^2}}$.At $x = 1$:$\left(\frac{dy}{dx}ight)_{x=1} = \frac{1}{\sqrt{1 + 1^2}} = \frac{1}{\sqrt{2}}$.

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

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