Exercise
Find the derivative of the following function:
f ( x ) = x + x 3 f(x)=\sqrt[3]{x+\sqrt{x}} f ( x ) = 3 x + x
Final Answer
Show 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 + x ) 1 3 f(x)={(x+\sqrt{x})}^{\frac{1}{3}} f ( x ) = ( x + x ) 3 1
Using Derivative formulas
f ′ ( x ) = 1 3 ( x + x ) − 2 3 ⋅ ( 1 + 1 2 x ) = f'(x)=\frac{1}{3}{(x+\sqrt{x})}^{-\frac{2}{3}}\cdot (1+ \frac{1}{2\sqrt{x}})= f ′ ( x ) = 3 1 ( x + x ) − 3 2 ⋅ ( 1 + 2 x 1 ) =
= 1 3 ( x + x ) 2 3 ⋅ 2 x + 1 2 x = =\frac{1}{3\sqrt[3]{{(x+\sqrt{x})}^2}}\cdot \frac{2\sqrt{x}+1}{2\sqrt{x}}= = 3 3 ( x + x ) 2 1 ⋅ 2 x 2 x + 1 =
= 2 x + 1 6 x ( x + x ) 3 2 =\frac{2\sqrt{x}+1}{6\sqrt{x}\sqrt[3]{(x+\sqrt{x})}^2} = 6 x 3 ( x + x ) 2 2 x + 1
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