#### please solve rd sharma class 12 chapter 16 Increasing and Decreasing Functions exercise Multiple choice question , question 3 maths textbook solution

Hint: If $f(x)$  is decreasing function ${f}'(x)<0$

Given:$f(x)=x^{x}$

Explanation: It is given that

$f(x)=x^{x}$

Taking log on both sides,

\begin{aligned} &\log f(x)=\log x^{x} \\ &\log f(x)=x \log x \quad\left[\because \log a^{b}=b \log a\right] \end{aligned}

Differentiate w.r.t  x

\begin{aligned} &\frac{1}{f(x)} \cdot f^{\prime}(x)=x \cdot \frac{1}{x}+\log x .1 \end{aligned}              [$\because$ by using u.v. rule]

\begin{aligned} &\frac{f^{\prime}(x)}{f(x)}=1+\log x\\ &\therefore f^{\prime}(x)=x^{x}( 1+\log x) \quad\left[\because f(x)=x^{x}\right] \end{aligned}\begin{aligned} &\frac{f^{\prime}(x)}{f(x)}=1+\log x\\ &\therefore f^{\prime}(x)=x^{x}( 1+\log x) \quad\left[\because f(x)=x^{x}\right] \end{aligned}

$f(x)$  is decreasing function ${f}{}'(x)<0$

\begin{aligned} &\Rightarrow x^{x}(1+\log x)<0 \\ &\Rightarrow 1+\log x<0 \\ &\Rightarrow \log x<-1 \\ &\Rightarrow x1 \Rightarrow e^{-1}<1 \end{aligned}

Thus the function is decreasing on $\left ( 0,\frac{1}{e} \right )$