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

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(C) Given $y = \sec(	an^{-1} x)$.Using the chain rule,we differentiate with respect to $x$:$\frac{dy}{dx} = \sec(	an^{-1} x) \cdot 	an(	an^{-1} x)

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

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(C) Given $y = \sec(	an^{-1} x)$.Using the chain rule,we differentiate with respect to $x$:$\frac{dy}{dx} = \sec(	an^{-1} x) \cdot 	an(	an^{-1} x) \cdot \frac{d}{dx}(	an^{-1} x)$$\frac{dy}{dx} = \sec(	an^{-1} x) \cdot x \cdot \frac{1}{1+x^2}$Since $	an^{-1} x = 	heta$,then $	an 	heta = x$,which implies $\sec 	heta = \sqrt{1+x^2}$.Substituting this into the derivative:$\frac{dy}{dx} = \sqrt{1+x^2} \cdot x \cdot \frac{1}{1+x^2} = \frac{x}{\sqrt{1+x^2}}$Now,evaluating 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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