If y = log_e(log_e x) for x 1,then frac{d^2 | vedclass

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(C) Given $y = \log_e(\log_e x)$.First,find the first derivative $\frac{dy}{dx}$ using the chain rule:$\frac{dy}{dx} = \frac{1}{\log_e x} \cdot \frac{

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$-\frac{\log_e e x}{x^2 (\log_e x)^2}$

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(C) Given $y = \log_e(\log_e x)$.First,find the first derivative $\frac{dy}{dx}$ using the chain rule:$\frac{dy}{dx} = \frac{1}{\log_e x} \cdot \frac{d}{dx}(\log_e x) = \frac{1}{x \log_e x} = (x \log_e x)^{-1}$.Now,find the second derivative $\frac{d^2 y}{dx^2}$ using the power rule and product rule:$\frac{d^2 y}{dx^2} = -1(x \log_e x)^{-2} \cdot \frac{d}{dx}(x \log_e x)$.$\frac{d^2 y}{dx^2} = -\frac{1}{(x \log_e x)^2} \cdot [1 \cdot \log_e x + x \cdot \frac{1}{x}] = -\frac{\log_e x + 1}{x^2 (\log_e x)^2}$.Since $\log_e x + 1 = \log_e x + \log_e e = \log_e(ex)$,we have:$\frac{d^2 y}{dx^2} = -\frac{\log_e(ex)}{x^2 (\log_e x)^2}$.

If $y = \log_e(\log_e x)$ for $x > 1$,then $\frac{d^2 y}{dx^2} = $ . . . . . . .

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