Chapter 11 - Section 11.5 - Derivatives of Logarithmic and Exponential Functions - Exercises - Page 842: 15

$\ln x+1$

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