Answer
$y'=\frac{(x+1)(x-5)^{3}}{(x-3)^{8}}[\frac{4}{x+1}+\frac{3}{x-5}-\frac{8}{x-3}]$
Work Step by Step
Use logarithmic differentiation to find the derivative of the function.
$y=\frac{(x+1)(x-5)^{3}}{(x-3)^{8}}$
Taking logarthimic on both sides .
$lny=ln[\frac{(x+1)(x-5)^{3}}{(x-3)^{8}}]$
Use logarithmic properties $ln(xy)=lnx+lny$ , $ln(\frac{x}
{y})=lnx-lny$ and $ln(x^{y})=ylnx$.
$lny=4ln(x+1)+3ln(x-5)-8ln(x-3)$
Differentiate with respect to $x$.
$\frac{1}{y}\frac{dy}{dx}=\frac{4}{x+1}+\frac{3}{x-5}-\frac{8}{x-3}$
$y'=y[\frac{4}{x+1}+\frac{3}{x-5}-\frac{8}{x-3}]$
Hence,
$y'=\frac{(x+1)(x-5)^{3}}{(x-3)^{8}}[\frac{4}{x+1}+\frac{3}{x-5}-\frac{8}{x-3}]$
