Question

find the first and second derivative of the function.
g(r) = \sqrt{r}+\sqrt7{r}
g(r)=
g(r)=

Explanation:

Step1: Rewrite the function

Rewrite $G(r)=\sqrt{r}+\sqrt[7]{r}$ as $G(r)=r^{\frac{1}{2}}+r^{\frac{1}{7}}$.

Step2: Find the first - derivative

Use the power rule $\frac{d}{dr}(r^n)=nr^{n - 1}$.
$G'(r)=\frac{1}{2}r^{\frac{1}{2}-1}+\frac{1}{7}r^{\frac{1}{7}-1}=\frac{1}{2}r^{-\frac{1}{2}}+\frac{1}{7}r^{-\frac{6}{7}}$.

Step3: Find the second - derivative

Differentiate $G'(r)$ again using the power rule.
$G''(r)=\frac{1}{2}\times(-\frac{1}{2})r^{-\frac{1}{2}-1}+\frac{1}{7}\times(-\frac{6}{7})r^{-\frac{6}{7}-1}=-\frac{1}{4}r^{-\frac{3}{2}}-\frac{6}{49}r^{-\frac{13}{7}}$.

Answer:

$G'(r)=\frac{1}{2}r^{-\frac{1}{2}}+\frac{1}{7}r^{-\frac{6}{7}}$
$G''(r)=-\frac{1}{4}r^{-\frac{3}{2}}-\frac{6}{49}r^{-\frac{13}{7}}$