Answer
$y'=\sqrt xe^{x^{2}-x}(x+1)^{2/3}[\frac{1}{2x}+(2x-1)+\frac{2}{3(x+1)}]$
Work Step by Step
Use logarithmic differentiation to find the derivative of the function
$y=\sqrt xe^{x^{2}-x}(x+1)^{2/3}$
Take the log on both sides
$lny= ln[\sqrt xe^{x^{2}-x}(x+1)^{2/3}]$
Use logarithmic properties $ln(xy)=lnx+lny$ and $ln(x^{y})=ylnx$.
$lny=\frac{1}{2}lnx+({x^{2}-x})+\frac{2}{3}ln(x+1)$
Differentiate with respect to $x$.
$y'=y[\frac{1}{2x}+(2x-1)+\frac{2}{3(x+1)}]$
Hence, $y'=\sqrt xe^{x^{2}-x}(x+1)^{2/3}[\frac{1}{2x}+(2x-1)+\frac{2}{3(x+1)}]$
