Question

The function $x^x$ is increasing, when

• A. $x>\frac{1}{e}$
• B. $x<\frac{1}{e}$
• C. $x<0$
• D. $\forall x$

Answer 1

The correct option is A

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

Explanation for the correct option

Given function is $x^x$

Let $y=x^x$
$\begin{array}{rcl}y& =& e^{\ln \left(x^x\right)}\left[\because e^{\ln \left(a\right)}=a\right]\\ \frac{dy}{dx}& =& \frac{d}{dx}\left(e^{\ln \left(x^x\right)}\right)\\ & =& \frac{d}{dx}\left(e^{x\ln x}\right)\\ & =& e^{x\ln x}\left(\frac{dx}{dx}·\ln x+x·\frac{d}{dx}\ln x\right)\left[\because \frac{d}{dx}\left[f\left(g\left(x\right)\right)\right]=f\text{'}\left(g\left(x\right)\right).g\text{'}\left(x\right),\frac{d}{dx}\left(U·V\right)=U·\frac{dV}{dx}+V·\frac{dU}{dx}\right]\\ & =& e^{\ln \left(x^x\right)}\left(\ln x+x·\frac{1}{x}\right)\\ & =& x^x\left(\ln x+1\right)\end{array}$

A function is increasing if $f\text{'}\left(x\right)>0$. So,
$\begin{array}{ccc}& x^x\left(\ln x+1\right)>0& \\ \Rightarrow & \ln x+1>0& \left[\because x^x>0\right]\\ \Rightarrow & \ln x>-1& \\ \Rightarrow & x>\frac{1}{e}& \left[\because \ln \left(x\right)=y\Rightarrow x=e^y\right]\end{array}$

Thus, the function is increasing if $x>\frac{1}{e}$.

Hence, option(A) i.e. $x>\frac{1}{e}$ is correct.