Answer
$ \ln x(8x-1)+4x-1$
Work Step by Step
$u(x)=4x^{2}-x,\qquad u^{\prime}(x)=4(2x)-1=8x-1$
$v(x)=\displaystyle \ln x,\qquad v^{\prime}(x)=\frac{1}{x}$
$f(x)=u(x)v(x)$, so by the Product Rule,
$f^{\prime}(x)=u^{\prime}(x)v(x)+u(x)v^{\prime}(x)$
$=(8x-1)\displaystyle \cdot\ln x+(4x^{2}-x)\cdot\frac{1}{x}$
$= \displaystyle \ln x(8x-1)+\frac{x(4x-1)}{x}=$
$= \ln x(8x-1)+4x-1$
