# $f (x) = x^x$ has a stationary point at A. x = e B. $x=\frac{1}{e}$ C. x = 1 D. $x=\sqrt e$

### Answers (1)

Given equation is $f (x) = x^x$

Let $y= x^x$………(i)

Take logarithm on both side

$\log y=\log (x^x)$

⇒ log y=x log x

Apply first derivative and get

$\frac{\mathrm{d}(\log \mathrm{y})}{\mathrm{d} \mathrm{x}}=\frac{\mathrm{d}(\mathrm{x} \log \mathrm{x})}{\mathrm{d} \mathrm{x}}$
Apply product rule and get
$\frac{\mathrm{d}(\log \mathrm{y})}{\mathrm{dx}}=\mathrm{x} \frac{\mathrm{d}(\log \mathrm{x})}{\mathrm{d} \mathrm{x}}+\log \mathrm{x} \cdot \frac{\mathrm{d}(\mathrm{x})}{\mathrm{d} \mathrm{x}}$

Apply first derivative and get
$\\ \frac{1}{y} \cdot \frac{d y}{d x}=x \cdot \frac{1}{x}+\log x \cdot 1 \\\Rightarrow \frac{d y}{d x}=y(1+\log x)$
Substitute value of y from (i) and get
$\Rightarrow \frac{d y}{d x}=x^{x}(1+\log x)$

Now we find the critical point by equating (i) to 0 and get

$x^x (1 + \log x) = 0 \\ \Rightarrow 1 + \log x = 0 \quad \text{as } x^x \text{ cannot be equal to } 0$

$\\ { \Rightarrow log x=-1}\\ {But -1=log e\textsuperscript{-1}}\\ { \Rightarrow log x=log e\textsuperscript{-1}}\\$

Equate the terms and get

$x= e\textsuperscript{-1}$

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

Therefore f(x) has a stationary point at $X=\frac{1}{e}$

So, the correct answer is option B.

View full answer