fbpx
calculus online - Free exercises and solutions to help you succeed!

Calculating Derivative – Third root – Exercise 1106

Exercise

Find the derivative of the following function:

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

Final Answer


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}

Have a question? Found a mistake? – Write a comment below!
Was it helpful? You can buy me a cup of coffee here, which will make me very happy and will help me upload more solutions! 

Share with Friends

Leave a Reply