Question

$x^x$ has a stationary point at

• A. $x=e$
• B. $x=\frac{1}{e}$
• C. $x=1$
• D. $x=\sqrt{e}$

Answer 1

The correct option is B

$x=\frac{1}{e}$

Explanation for the correct option:

Step $1$: Finding the first derivative

Let's consider,

$y=x^x$

Take ${\log }_e$ on both sides, we get

${\log }_{e}y=x\left({\log }_{e}x\right),x>0$ $\left[\because \log \left(a^b\right)=b\log \left(a\right)\right]$

Differentiate with respect to $x$, we get

$\frac{1}{y}\frac{dy}{dx}=1+{\log }_{e}x$ $\left[\because \frac{d}{dx}\left(f\left(x\right)g\left(x\right)\right)=f\text{'}\left(x\right)g\left(x\right)+f\left(x\right)g\text{'}\left(x\right)\right]$

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

$\Rightarrow$ $\frac{dy}{dx}=x^x\left(1+{\log }_{e}x\right)$ $\left[\because y=x^x\right]$

Step $2$: Finding the stationary point

We know that, at a stationary point $\frac{dy}{dx}=0$.

So,

$x^x\left(1+{\log }_{e}x\right)=0$

$\Rightarrow$ $1+{\log }_{e}x=0$ $\left[\because x^{x}cannotbe0\right]$

$\Rightarrow$ ${\log }_{e}x=-1$

$\Rightarrow$ $x=e^{-1}$ $\left[if{\log }_a\left(m\right)=n,then{a}^n=m\right]$

$\Rightarrow$ $x=\frac{1}{e}$

The stationary point is $x=\frac{1}{e}$.

Hence, the correct option is $B)x=\frac{1}{e}$