Question

If $y=x^3\log {\log }_e(1+x)$, then $y\text{'}\text{'}\left(0\right)$ equals

• A. $0$
• B. $-1$
• C. $6{\log }_{e}2$
• D. $6$

Answer 1

The correct option is A

$0$

Explanation for the correct option.

Step 1: Find first derivative.

$\begin{array}{rcl}y& =& x^3\left[\log \left\{{\log }_e(1+x)\right\}\right]\\ y\text{'}& =& 3x^2\left[\log \left\{\log (1+x)\right\}\right]+x^3\left[\frac{1}{\log (1+x)}·\frac{1}{1+x}·1\right]\\ & =& x^2\left[3\log \left\{\log (1+x)\right\}\right]+\left[\frac{x}{\left(1+x\right)\log (1+x)}\right]\end{array}$

Step 2: Find second derivative.

$y\text{'}\text{'}=2x\begin{array}{rcl}& & \left[3\log \left\{\log (1+x)\right\}\right]\end{array}\begin{array}{rcl}& & +\end{array}\begin{array}{rcl}& & \left[\frac{x}{\left(1+x\right)\log (1+x)}\right]\end{array}\begin{array}{rcl}& & +\end{array}{x}^2\begin{array}{rcl}& & \left[\frac{3}{\log (1+x)(1+x)}\right]\end{array}\begin{array}{rcl}& & +\end{array}\begin{array}{rcl}& & \left[\frac{\left(1+x\right)\log (1+x)-x\left\{{\displaystyle \frac{1+x+\log (1+x)}{1+x}}\right\}}{{\left\{\left(1+x\right)\log (1+x)\right\}}^2}\right]\end{array}$

Now,

$\begin{array}{rcl}y\text{'}\text{'}\left(0\right)& =& 2\left(0\right)\begin{array}{l}\left[3\log \left\{\log (1+0)\right\}\right]\end{array}\begin{array}{l}+\left[\frac{0}{\left(1+0\right)\log (1+0)}\right]\end{array}+0^2\begin{array}{l}\left[\frac{3}{\log (1+0)(1+0)}\right]+\end{array}\begin{array}{l}\left[\frac{\left(1+0\right)\log (1+0)-0\left\{{\displaystyle \frac{1+0+\log (1+0)}{1+0}}\right\}}{{\left\{\left(1+0\right)\log (1+0)\right\}}^2}\right]\end{array}\\ & =& 0\end{array}$

Hence, option A is correct.