Find the second order derivative of the funct | vedclass

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Question

(A) Let $y = \log (\log x)$.Differentiating with respect to $x$,we get:$\frac{dy}{dx} = \frac{d}{dx}[\log (\log x)] = \frac{1}{\log x} \cdot \frac{d}{

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$\frac{-(1+\log x)}{(x \log x)^2}$

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(A) Let $y = \log (\log x)$.Differentiating with respect to $x$,we get:$\frac{dy}{dx} = \frac{d}{dx}[\log (\log x)] = \frac{1}{\log x} \cdot \frac{d}{dx}(\log x) = \frac{1}{x \log x} = (x \log x)^{-1}$.Now,differentiating again with respect to $x$ to find the second order derivative:$\frac{d^2y}{dx^2} = \frac{d}{dx}[(x \log x)^{-1}] = -1 \cdot (x \log x)^{-2} \cdot \frac{d}{dx}(x \log x)$.Using the product rule for $\frac{d}{dx}(x \log x)$:$\frac{d}{dx}(x \log x) = x \cdot \frac{d}{dx}(\log x) + \log x \cdot \frac{d}{dx}(x) = x \cdot \frac{1}{x} + \log x \cdot 1 = 1 + \log x$.Substituting this back:$\frac{d^2y}{dx^2} = \frac{-1}{(x \log x)^2} \cdot (1 + \log x) = \frac{-(1 + \log x)}{(x \log x)^2}$.

Find the second order derivative of the function $\log (\log x)$.

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