If y = log(sec(tan^{-1} x)) for x 0,then th | vedclass

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

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$\frac{1}{2}$

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(C) Given $y = \log(\sec(	an^{-1} x))$.Using the chain rule,$\frac{dy}{dx} = \frac{1}{\sec(	an^{-1} x)} \cdot \frac{d}{dx}(\sec(	an^{-1} x))$.Recall that $\frac{d}{dx}(\sec u) = \sec u 	an u \cdot \frac{du}{dx}$.Let $u = 	an^{-1} x$,then $\frac{du}{dx} = \frac{1}{1+x^2}$.So,$\frac{dy}{dx} = \frac{1}{\sec(	an^{-1} x)} \cdot \sec(	an^{-1} x) 	an(	an^{-1} x) \cdot \frac{1}{1+x^2}$.This simplifies to $\frac{dy}{dx} = 	an(	an^{-1} x) \cdot \frac{1}{1+x^2} = x \cdot \frac{1}{1+x^2} = \frac{x}{1+x^2}$.At $x = 1$,$\frac{dy}{dx} = \frac{1}{1+(1)^2} = \frac{1}{2}$.

If $y = \log(\sec(\tan^{-1} x))$ for $x > 0$,then the value of $\frac{dy}{dx}$ at $x = 1$ is:

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