If y = x^3 log(log_e(1 + x)),then y''(0) equa | vedclass

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(A) Given $y = x^3 \log(\log_e(1 + x))$.First,find the first derivative $y'$ using the product rule:$y' = 3x^2 \log(\log_e(1 + x)) + x^3 \cdot \frac{1

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(A) Given $y = x^3 \log(\log_e(1 + x))$.First,find the first derivative $y'$ using the product rule:$y' = 3x^2 \log(\log_e(1 + x)) + x^3 \cdot \frac{1}{\log_e(1 + x)} \cdot \frac{1}{1 + x}$$y' = 3x^2 \log(\log_e(1 + x)) + \frac{x^3}{(1 + x) \log_e(1 + x)}$Now,find the second derivative $y''$:$y'' = \frac{d}{dx} [3x^2 \log(\log_e(1 + x))] + \frac{d}{dx} \left[ \frac{x^3}{(1 + x) \log_e(1 + x)} ight]$As $x 	o 0$,the term $3x^2 \log(\log_e(1 + x))$ involves $x^2 \log(\log_e(1+x))$. Since $\log_e(1+x) \approx x$ for small $x$,$\log(\log_e(1+x)) \approx \log(x)$,and $x^2 \log(x) 	o 0$ as $x 	o 0$.For the second term $\frac{x^3}{(1 + x) \log_e(1 + x)}$,using the expansion $\log_e(1+x) = x - \frac{x^2}{2} + \dots$,we have:$\frac{x^3}{(1 + x)(x - \frac{x^2}{2})} = \frac{x^3}{x(1 + x)(1 - \frac{x}{2})} = \frac{x^2}{(1 + x)(1 - \frac{x}{2})} 	o 0$ as $x 	o 0$.Evaluating the derivatives at $x=0$ yields $y''(0) = 0$.

If $y = x^3 \log(\log_e(1 + x))$,then $y''(0)$ equals

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