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

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

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

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

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\frac{d}{d x} (3 x + 8)

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

Solution steps

\frac{d}{d x} (3 x + 8)

Apply the chain rule:

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

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

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

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

Simplify

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

\int cos^{2} (x t a) n^{3} x d x

3 u + 5 u

y^{^{'}} = 4 - 9^{2} - 6 x^{5}

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\int_{0}^{6} 36 - x^{2} d x

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