Question

Find the second order derivatives of $x^3.\log x$

Answer 1

Let $y=x^3\log x$
$\Rightarrow \frac{dy}{dx}=\log x\frac{d}{dx}\left(x^3\right)+x^3.\frac{d}{dx}\left(\log x\right)$
$=\log x.3x^2+x^3.\frac{1}{x}=\log x.3x^2+x^2$
$=x^2(1+3\log x)$
$\therefore$ $\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left[x^2\right(1+3\log x\left)\right]$
$=(1+3\log x).\frac{d}{dx}\left(x^2\right)+x^2\frac{d}{dx}(1+3\log x)$
$=(1+3\log x).2x+x^2\frac{3}{x}$
$=2x+6x\log x+3x=5x+6x\log x=x(5+6\log x)$