What is the derivative of y=\sqrt[3]{1+8x} ?

The derivative of y=\sqrt[3]{1+8x} is 8/(3(1+8x)^{2/3)}

What is the first derivative of y=\sqrt[3]{1+8x} ?

The first derivative of y=\sqrt[3]{1+8x} is 8/(3(1+8x)^{2/3)}

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derivative of

\frac{8}{3 ( 1 + 8 x ) ^{\frac{2}{3}}}

Solution steps

Apply the chain rule:

\frac{1}{3 ( 1 + 8 x ) ^{\frac{2}{3}}} \frac{d}{d x} (1 + 8 x)

= \frac{1}{3 ( 1 + 8 x ) ^{\frac{2}{3}}} \frac{d}{d x} (1 + 8 x)

\frac{d}{d x} (1 + 8 x) = 8

= \frac{1}{3 ( 1 + 8 x ) ^{\frac{2}{3}}} \cdot 8

Simplify

\frac{1}{3 ( 1 + 8 x ) ^{\frac{2}{3}}} \cdot 8 : \frac{8}{3 ( 1 + 8 x ) ^{\frac{2}{3}}}

= \frac{8}{3 ( 1 + 8 x ) ^{\frac{2}{3}}}

What is the derivative of y=\sqrt[3]{1+8x} ?
The derivative of y=\sqrt[3]{1+8x} is 8/(3(1+8x)^{2/3)}
What is the first derivative of y=\sqrt[3]{1+8x} ?
The first derivative of y=\sqrt[3]{1+8x} is 8/(3(1+8x)^{2/3)}