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