find the first and second derivative of the function. g(r) = √r + ⁴√r g(r) = g(r) =

Explanation:

Step1: Rewrite the function

Rewrite $G(r)=\sqrt{r}+\sqrt[4]{r}$ as $G(r)=r^{\frac{1}{2}}+r^{\frac{1}{4}}$.

Step2: Find the first - derivative

Use the power rule $(x^n)' = nx^{n - 1}$. For $y = r^{\frac{1}{2}}$, $y'=\frac{1}{2}r^{\frac{1}{2}-1}=\frac{1}{2}r^{-\frac{1}{2}}$, and for $y = r^{\frac{1}{4}}$, $y'=\frac{1}{4}r^{\frac{1}{4}-1}=\frac{1}{4}r^{-\frac{3}{4}}$. So $G'(r)=\frac{1}{2}r^{-\frac{1}{2}}+\frac{1}{4}r^{-\frac{3}{4}}$.

Step3: Find the second - derivative

Differentiate $G'(r)$ again using the power rule. For $y=\frac{1}{2}r^{-\frac{1}{2}}$, $y'=\frac{1}{2}\times(-\frac{1}{2})r^{-\frac{1}{2}-1}=-\frac{1}{4}r^{-\frac{3}{2}}$, and for $y = \frac{1}{4}r^{-\frac{3}{4}}$, $y'=\frac{1}{4}\times(-\frac{3}{4})r^{-\frac{3}{4}-1}=-\frac{3}{16}r^{-\frac{7}{4}}$. So $G''(r)=-\frac{1}{4}r^{-\frac{3}{2}}-\frac{3}{16}r^{-\frac{7}{4}}$.

Answer:

$G'(r)=\frac{1}{2}r^{-\frac{1}{2}}+\frac{1}{4}r^{-\frac{3}{4}}$ $G''(r)=-\frac{1}{4}r^{-\frac{3}{2}}-\frac{3}{16}r^{-\frac{7}{4}}$