Answer
$y'=\frac{(x^{3}+1)^{4}sin^{2}x}{x^{1/3}}[\frac{12x^{2}}{(x^{3}+1)}+2cotx-\frac{1}{3x}]$
Work Step by Step
Use logarithmic differentiation to find the derivative of the function.
$y=\frac{(x^{3}+1)^{4}sin^{2}x}{x^{1/3}}$
Taking logarthimic on both sides .
$lny=ln[\frac{(x^{3}+1)^{4}sin^{2}x}{x^{1/3}}]$
Use logarithmic properties $ln(xy)=lnx+lny$ , $ln(\frac{x}
{y})=lnx-lny$ and $ln(x^{y})=ylnx$.
$lny=4ln{(x^{3}+1)}+2lnsinx-\frac{1}{3}lnx$
Differentiate with respect to $x$
$y'=y[\frac{12x^{2}}{(x^{3}+1)}+2cotx-\frac{1}{3x}$
Hence,
$y'=\frac{(x^{3}+1)^{4}sin^{2}x}{x^{1/3}}[\frac{12x^{2}}{(x^{3}+1)}+2cotx-\frac{1}{3x}]$
