Answer
$$\dfrac{dy}{dx}=(x^{x-1} 3^x)[ \dfrac{(x-1)}{x}+\dfrac{\ln x}{x-1}+\ln (3)]$$
Work Step by Step
We have: $y=x^{x-1} 3^x$
or, $\ln y=\ln [x^{x-1} 3^x]$
We differentiate both sides with respect to $x$.
$\ln (y) =(x-1) \ln x +x\ln (3) $
This implies that $\dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{(x-1)}{x}+\dfrac{\ln x}{x-1}+\ln (3)$
Therefore, $\dfrac{dy}{dx}=y \times \dfrac{(x-1)}{x}+\dfrac{\ln x}{x-1}+\ln (3)\\ \dfrac{dy}{dx}=(x^{x-1} 3^x)[ \dfrac{(x-1)}{x}+\dfrac{\ln x}{x-1}+\ln (3)]$
