A calculus problem by Md Zuhair

$\begin{aligned} y & = x^x \\ \ln y & = x \ln x \\ \frac 1y \cdot \frac {dy}{dx} & = \ln x + 1 & \small \color{#3D99F6} \text{When } x= 1, \ y = 1 \\ \frac 11 \cdot \frac {dy}{dx} \bigg|_{x=1} & = \ln 1 + 1 \\ \implies \frac {dy}{dx} \bigg|_{x=1} & = \boxed{1} \end{aligned}$

If $y$ = $x^x$ , then we need to use the method of logarithmic differentiation:

$ln$ $y$ = $ln$ $x^x$ = $xln$ $x$ .

$\frac{d}{dx}$ $ln$ $y$ = $\frac{d}{dx}$ $xln$ $x$ .

$\frac{dy}{dx}$ $\frac{1}{y}$ = $ln$ $x$ + $1$ .

If we then multiply both sides by $y$ to get rid of the $\frac{1}{y}$ term on the left side of the equation, we will get:

$\frac{dy}{dx}$ = $x^x(ln$ $x$ + $1)$ .

Finally, we input $1$ into the equation to get:

$\frac{dy}{dx}$ at $x$ = $1$ is equal to:

$1^1(ln$ $1$ + $1)$ = $1$ .