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If y=log⁡(log⁡x)y = \log(\log x), then d2ydx2\frac{d^2y}{dx^2} is equal to: (Consider log⁡ex=log⁡x\log_e x = \log x)

Solution

✅ Correct Option: 2

dydx=1xlog⁡x=(xlog⁡x)−1\frac{dy}{dx}=\frac{1}{x\log x}=(x\log x)^{-1}. Differentiating again, d2ydx2=−(xlog⁡x)−2⋅ddx(xlog⁡x)\frac{d^2y}{dx^2}=-(x\log x)^{-2}\cdot\frac{d}{dx}(x\log x). Since ddx(xlog⁡x)=1+log⁡x\frac{d}{dx}(x\log x)=1+\log x, we get −1+log⁡x(xlog⁡x)2-\frac{1+\log x}{(x\log x)^2}.

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