Question

If $y=x^x,\text{then prove that}\frac{d^{2}y}{d{x}^2}-\frac{1}{y}{\left(\frac{dy}{dx}\right)}^2-\frac{y}{x}=0$.

Answer 1

We have,

$y=x^x$

On taking logarithm both sides, we get

$\ln y=\ln \left(x^x\right)$ $............\left(1\right)$

$\ln y=x\ln x$

On differentiating w.r.t $x$, we get

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

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

$\frac{dy}{dx}=y(1+\ln x)$ $............\left(2\right)$

On differentiating w.r.t $x$, we get

$\frac{d^{2}y}{d{x}^2}=\frac{dy}{dx}(1+\ln x)+y(0+\frac{1}{x})$

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

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

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

$\frac{d^{2}y}{d{x}^2}-\frac{1}{y}{\left(\frac{dy}{dx}\right)}^2-\frac{y}{x}=0$

Hence, proved.