# Calculating Derivative – Third root – Exercise 1106

Exercise

Find the derivative of the following function:

$$f(x)=\sqrt[3]{x+\sqrt{x}}$$

$$f'(x)=\frac{2\sqrt{x}+1}{6\sqrt{x}\sqrt[3]{(x+\sqrt{x})}^2}$$

Solution

First, we simplify the function:

$$f(x)={(x+\sqrt{x})}^{\frac{1}{3}}$$

Using Derivative formulas, we get the derivative:

$$f'(x)=\frac{1}{3}{(x+\sqrt{x})}^{-\frac{2}{3}}\cdot (1+ \frac{1}{2\sqrt{x}})=$$

$$=\frac{1}{3\sqrt[3]{{(x+\sqrt{x})}^2}}\cdot \frac{2\sqrt{x}+1}{2\sqrt{x}}=$$

$$=\frac{2\sqrt{x}+1}{6\sqrt{x}\sqrt[3]{(x+\sqrt{x})}^2}$$

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