Question

Find the second order derivative of the function $y=x^3\log x$

Answer 1

Finding $\frac{dy}{dx}$
Let $y=x^3\log x$
Differentiating both sides w.r.t. $x$ we get
$\frac{dy}{dx}=\frac{d\left(x^3\log x\right)}{dx}$
Using Product Rule, $(uv{)}^{\prime }=u^{\prime }v+v^{\prime }u$
$\frac{dy}{dx}=\frac{d\left(x^3\right)}{dx}.\log x+\frac{d\left(\log x\right)}{dx}.x^3$
$\Rightarrow \frac{dy}{dx}=3x^2\log x+\frac{1}{x}.x^3$
$\Rightarrow \frac{dy}{dx}=3x^2\log x+x^2\cdots \left(i\right)$

Finding $\frac{d^{2}y}{d{x}^2}$
Again, differentiating both sides of $\left(i\right)$ w.r.t. $x$, we get
$\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d(3x^2\log x+x^2)}{dx}$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{d\left(3x^2\log x\right)}{dx}+\frac{d\left(x^2\right)}{dx}$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=3.\frac{d(x^2.\log x)}{dx}+2x$
Using Product Rule, $(uv{)}^{\prime }=u^{\prime }v+v^{\prime }u$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=3(\frac{d(x{)}^2}{dx}.\log x+\frac{d\left(\log x\right)}{dx}.x^2)+2x$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=3(2x\log x+\frac{1}{x}.x^2)+2x$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=3(2x\log x+x)+2x$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=6x\log x+3x+2x$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=6x\log x+5x$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=x(6\log x+5)$