Question

State whether the statement is true/false.

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

• A. True
• B. False

Answer 1

The correct option is B False

$y=x^x$

Taking log on both sides,

$\log y=x\log x$

Differentiate w.r.t $x,$ we have

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

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

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

Again differentiating w.r.t. $x,$

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

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

We have given,

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

L.H.S $=\frac{d^{2}y}{d{x}^2}-\frac{1}{y}{\left(\frac{dy}{dx}\right)}^2\frac{y}{x}$

$=\frac{y}{x}+y(1+\log x{)}^2-\frac{1}{y}\times y^2(1+\log x{)}^2-\frac{y}{x}$ [ From ( 1 ) and ( 2 ) ]

$=0$

$\therefore$ $L.H.S\ne R.H.S$

$\therefore$ $\frac{d^{2}y}{d{x}^2}-\frac{1}{y}{\left(\frac{dy}{dx}\right)}^2\frac{y}{x}=1$ is false.