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

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(A) Given that,$y = \log(\log x) \quad (1)$Differentiating Eq. $(1)$ with respect to $x$,we get:$\frac{dy}{dx} = \frac{1}{\log x} \cdot \frac{d}{dx}(\

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

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(A) Given that,$y = \log(\log x) \quad (1)$Differentiating Eq. $(1)$ with respect to $x$,we get:$\frac{dy}{dx} = \frac{1}{\log x} \cdot \frac{d}{dx}(\log x) = \frac{1}{\log x} \cdot \frac{1}{x} = \frac{1}{x \log x} \quad (2)$Now,differentiating Eq. $(2)$ with respect to $x$ using the quotient rule $\frac{d}{dx}(\frac{u}{v}) = \frac{v u' - u v'}{v^2}$:$\frac{d^2y}{dx^2} = \frac{d}{dx}((x \log x)^{-1}) = -1(x \log x)^{-2} \cdot \frac{d}{dx}(x \log x)$$\frac{d^2y}{dx^2} = -\frac{1}{(x \log x)^2} \cdot [x \cdot \frac{1}{x} + \log x \cdot 1]$$\frac{d^2y}{dx^2} = -\frac{1 + \log x}{(x \log x)^2}$

If $y = \log(\log x)$,then $\frac{d^2y}{dx^2}$ is equal to

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