Chapter 15 - Differentiation in Several Variables - 15.6 The Chain Rule - Exercises - Page 809: 36

$$\frac{1}{r^2}\mathbf{r}$$

Since $$F(r)=\ln r,\ \ \ \ \ \ \ \ F'(r) = \frac{ 1}{r } $$ Then \begin{align*} \nabla f &=F^{\prime}(r) e_{\mathbf{r}}\\ &= \frac{1}{r}e_{\mathbf{r}}\\ &= \frac{1}{r}\frac{\mathbf{r}}{r}\\ &=\frac{1}{r^2}\mathbf{r} \end{align*}