Question

find the first and second derivative of the function.
g(r) = \sqrt{r}+\sqrt6{r}
g(r) =
g(r) =

Explanation:

Step1: Rewrite the function

Rewrite $G(r)=\sqrt{r}+\sqrt[6]{r}$ as $G(r)=r^{\frac{1}{2}}+r^{\frac{1}{6}}$.

Step2: Find the first - derivative

Use the power rule $\frac{d}{dr}(r^n)=nr^{n - 1}$.
$G'(r)=\frac{1}{2}r^{\frac{1}{2}-1}+\frac{1}{6}r^{\frac{1}{6}-1}=\frac{1}{2}r^{-\frac{1}{2}}+\frac{1}{6}r^{-\frac{5}{6}}$.

Step3: Find the second - derivative

Differentiate $G'(r)$ using the power rule again.
$G''(r)=\frac{1}{2}\times(-\frac{1}{2})r^{-\frac{1}{2}-1}+\frac{1}{6}\times(-\frac{5}{6})r^{-\frac{5}{6}-1}=-\frac{1}{4}r^{-\frac{3}{2}}-\frac{5}{36}r^{-\frac{11}{6}}$.

Answer:

$G'(r)=\frac{1}{2}r^{-\frac{1}{2}}+\frac{1}{6}r^{-\frac{5}{6}}$
$G''(r)=-\frac{1}{4}r^{-\frac{3}{2}}-\frac{5}{36}r^{-\frac{11}{6}}$