Answer
$0.5(x+1)^{-0.5}\displaystyle \cdot\ln x+\frac{(x+1)^{0.5}}{x}$
Work Step by Step
$u(x)=(x+1)^{0.5},\qquad $so, by the Chain Rule,
$u^{\prime}(x)=0.5(x+1)^{-0.5}\cdot 1=0.5(x+1)^{-0.5}$
$v(x)=\displaystyle \ln x,\qquad v^{\prime}(x)=\frac{1}{x}$
$f(x)=u(x)v(x)$, so by the Product Rule,
$f^{\prime}(x)=u^{\prime}(x)v(x)+u(x)v^{\prime}(x)$
$=0.5(x+1)^{-0.5}\displaystyle \cdot\ln x+(x+1)^{0.5}\cdot\frac{1}{x}$
