## Calculus: Early Transcendentals 8th Edition

$y'=\sqrt xe^{x^{2}-x}(x+1)^{2/3}[\frac{1}{2x}+(2x-1)+\frac{2}{3(x+1)}]$
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, $y'=\sqrt xe^{x^{2}-x}(x+1)^{2/3}[\frac{1}{2x}+(2x-1)+\frac{2}{3(x+1)}]$