$\frac{dy}{dx}$ নির্ণয় কর $( y=x^{\ln x} \quad$)

Solve: Let, $y=x^{\ln x} \quad$
$\begin{array}{l} \therefore \quad \frac{d y}{d x}=x^{\ln x}\left[\ln x \frac{d}{d x}(\ln x)+\ln x \frac{d}{d x}(\ln x)\right] \\ {\left[\because \frac{d}{d x}\left(u^{v}\right)=u^{v}\left[v \frac{d}{d x}(\ln u)+\ln u \frac{d v}{d x}\right]\right.} \\ =x^{\ln x}\left[2 \ln x \frac{1}{x}\right]=\frac{2 \ln x}{x} x^{\ln x} \\ \text { So, } \frac{d}{d x}\left(x^{\ln x}\right)=\frac{2 \ln x}{x} x^{\ln x} \end{array}$