If y = tan(log x),then frac{d^2 y}{d x^2} = | vedclass

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(D) Given $y = 	an(\log x)$.First,differentiate with respect to $x$ using the chain rule:$\frac{dy}{dx} = \sec^2(\log x) \cdot \frac{1}{x} = \frac{\s

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$\frac{\sec^2(\log x)[2 	an(\log x) - 1]}{x^2}$

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(D) Given $y = 	an(\log x)$.First,differentiate with respect to $x$ using the chain rule:$\frac{dy}{dx} = \sec^2(\log x) \cdot \frac{1}{x} = \frac{\sec^2(\log x)}{x}$.Now,differentiate again with respect to $x$ using the quotient rule $\left( \frac{u}{v} ight)' = \frac{u'v - uv'}{v^2}$:$\frac{d^2y}{dx^2} = \frac{x \cdot \frac{d}{dx}(\sec^2(\log x)) - \sec^2(\log x) \cdot 1}{x^2}$.Using the chain rule for $\frac{d}{dx}(\sec^2(\log x)) = 2 \sec(\log x) \cdot \sec(\log x) 	an(\log x) \cdot \frac{1}{x} = \frac{2 \sec^2(\log x) 	an(\log x)}{x}$.Substituting this back:$\frac{d^2y}{dx^2} = \frac{x \cdot \frac{2 \sec^2(\log x) 	an(\log x)}{x} - \sec^2(\log x)}{x^2}$.$\frac{d^2y}{dx^2} = \frac{2 \sec^2(\log x) 	an(\log x) - \sec^2(\log x)}{x^2}$.Factoring out $\sec^2(\log x)$:$\frac{d^2y}{dx^2} = \frac{\sec^2(\log x)[2 	an(\log x) - 1]}{x^2}$.

If $y = \tan(\log x)$,then $\frac{d^2 y}{d x^2} =$

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