Question

If $y=x^x$, find $\frac{d^{2}y}{d{x}^2}$.

Answer 1

$y=x^x$

$\log y=x\log x$

Differentiating w.r.t. x, we get,

$\frac{1}{y}\frac{dy}{dx}=1\times \log x+x\times \frac{1}{x}$

$\frac{dy}{dx}=y(1+\log x)$

Differentiating both sides w.r.t. x, we get,

$\frac{d^{2}y}{d{x}^2}=\frac{dy}{dx}(1+\log x)+y\frac{d}{dx}(1+\log x)$

$\frac{d^{2}y}{d{x}^2}=\frac{dy}{dx}(1+\log x)+y\times \frac{1}{x}=y(1+\log x{)}^2+\frac{y}{x}$

$\frac{d^{2}y}{d{x}^2}=x^x\left[\right(1+\log x{)}^2+\frac{1}{x}]$