APPLICATION OF DERIVATIVES — Maths STD 12 Science — Question

2. $\text{x}=\frac{1}{\text{e}}$

Solution:

We have, f(x) = x^x

Let us suppose y = x^x

Taking logarithm on both sides, we get

$\log\text{y}=\text{x}\log\text{x}$

$\therefore\ \frac{1}{\text{y}}\cdot\frac{\text{dy}}{\text{dx}}=\text{x}\cdot\frac{1}{\text{x}}+\log\text{x}\cdot1$ $\big[\because(\text{fg})'=\text{fg}'+\text{gf}'\big]$

$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=(1+\log\text{x})\cdot\text{x}^\text{x}$

Find the critical points by equating $\frac{\text{dy}}{\text{dx}}$ to 0.

$\therefore\ \frac{\text{dy}}{\text{dx}}=0$

$\Rightarrow\ (1+\log\text{x})\text{x}^\text{x}=0$

$\Rightarrow\ \log\text{x}=-1$ as $\text{x}^\text{x}\neq0$

$\Rightarrow\ \log\text{x}=\log\text{e}^{-1}$

$\Rightarrow\ \text{x}=\text{e}^{-1}$

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

Hence, f(x) has a stationary point at $\text{x}=\frac{1}{\text{e}}.$