Answer
- First derivative: $$G'(r)=\frac{1}{2\sqrt r}+\frac{1}{3\sqrt[3]{r^2}}$$
- Second derivative: $$G''(r)=-\frac{1}{4\sqrt{r^3}}-\frac{2}{9\sqrt[3]{r^5}}$$
Work Step by Step
$$G(r)=\sqrt r+\sqrt[3]r$$$$G(r)=r^{1/2}+r^{1/3}$$
- First derivative: $$G'(r)=\frac{d}{dr}(r^{1/2})+\frac{d}{dr}(r^{1/3})$$
$$G'(r)=\frac{1}{2}r^{-1/2}+\frac{1}{3}r^{-2/3}$$
$$G'(r)=\frac{1}{2r^{1/2}}+\frac{1}{3r^{2/3}}$$
$$G'(r)=\frac{1}{2\sqrt r}+\frac{1}{3\sqrt[3]{r^2}}$$
- Second derivative: $$G''(r)=\frac{1}{2}\frac{d}{dr}(r^{-1/2})+\frac{1}{3}\frac{d}{dr}(r^{-2/3})$$
$$G''(r)=\frac{1}{2}\times(-\frac{1}{2})r^{-3/2}+\frac{1}{3}\times(-\frac{2}{3})r^{-5/3}$$
$$G''(r)=-\frac{1}{4}\times\frac{1}{\sqrt{r^3}}-\frac{2}{9}\times\frac{1}{\sqrt[3]{r^5}}$$
$$G''(r)=-\frac{1}{4\sqrt{r^3}}-\frac{2}{9\sqrt[3]{r^5}}$$
