Answer
$10x(x^{2}+1)^{4}\displaystyle \cdot\ln x+\frac{(x^{2}+1)^{5}}{x}$
Work Step by Step
$u(x)=(x^{2}+1)^{5},\qquad $so, by the Chain Rule,
$u^{\prime}(x)=5(x^{2}+1)^{4}\cdot 2x=10x(x^{2}+1)^{4}$
$v(x)=\displaystyle \ln x,\qquad v^{\prime}(x)=\frac{1}{x}$
$f(x)=u(x)v(x)$, so by the Product Rule,
$f^{\prime}(x)=u^{\prime}(x)v(x)+u(x)v^{\prime}(x)$
$=10x(x^{2}+1)^{4}\displaystyle \cdot\ln x+(x^{2}+1)^{5}\cdot\frac{1}{x}$
