What is the derivative of \sqrt[3]{3x+6} ?

The derivative of \sqrt[3]{3x+6} is 1/((3x+6)^{2/3)}

What is the first derivative of \sqrt[3]{3x+6} ?

The first derivative of \sqrt[3]{3x+6} is 1/((3x+6)^{2/3)}

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\frac{1}{( 3 x + 6 ) ^{\frac{2}{3}}}

Solution steps

Apply the chain rule:

\frac{1}{3 ( 3 x + 6 ) ^{\frac{2}{3}}} \frac{d}{d x} (3 x + 6)

= \frac{1}{3 ( 3 x + 6 ) ^{\frac{2}{3}}} \frac{d}{d x} (3 x + 6)

\frac{d}{d x} (3 x + 6) = 3

= \frac{1}{3 ( 3 x + 6 ) ^{\frac{2}{3}}} \cdot 3

Simplify

\frac{1}{3 ( 3 x + 6 ) ^{\frac{2}{3}}} \cdot 3 : \frac{1}{( 3 x + 6 ) ^{\frac{2}{3}}}

= \frac{1}{( 3 x + 6 ) ^{\frac{2}{3}}}

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What is the derivative of \sqrt[3]{3x+6} ?
The derivative of \sqrt[3]{3x+6} is 1/((3x+6)^{2/3)}
What is the first derivative of \sqrt[3]{3x+6} ?
The first derivative of \sqrt[3]{3x+6} is 1/((3x+6)^{2/3)}