What is the derivative of f(x)=\sqrt[3]{x^6+8} ?

The derivative of f(x)=\sqrt[3]{x^6+8} is (2x^5)/((x^6+8)^{2/3)}

What is the first derivative of f(x)=\sqrt[3]{x^6+8} ?

The first derivative of f(x)=\sqrt[3]{x^6+8} is (2x^5)/((x^6+8)^{2/3)}

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derivative of

\frac{2 x ^{5}}{( x ^{6} + 8 ) ^{\frac{2}{3}}}

Solution steps

Apply the chain rule:

\frac{1}{3 ( x ^{6} + 8 ) ^{\frac{2}{3}}} \frac{d}{d x} (x^{6} + 8)

= \frac{1}{3 ( x ^{6} + 8 ) ^{\frac{2}{3}}} \frac{d}{d x} (x^{6} + 8)

\frac{d}{d x} (x^{6} + 8) = 6 x^{5}

= \frac{1}{3 ( x ^{6} + 8 ) ^{\frac{2}{3}}} \cdot 6 x^{5}

Simplify

\frac{1}{3 ( x ^{6} + 8 ) ^{\frac{2}{3}}} \cdot 6 x^{5} : \frac{2 x ^{5}}{( x ^{6} + 8 ) ^{\frac{2}{3}}}

= \frac{2 x ^{5}}{( x ^{6} + 8 ) ^{\frac{2}{3}}}

\int_{0}^{1} \frac{33}{x ^{5}} d x

\int e^{- (s - a) t} d t

(3 + x^{5})^{\frac{2}{3}}

\frac{d}{d x} (π \cdot x)

\frac{d}{d t} (cos (t^{2}))

What is the derivative of f(x)=\sqrt[3]{x^6+8} ?
The derivative of f(x)=\sqrt[3]{x^6+8} is (2x^5)/((x^6+8)^{2/3)}
What is the first derivative of f(x)=\sqrt[3]{x^6+8} ?
The first derivative of f(x)=\sqrt[3]{x^6+8} is (2x^5)/((x^6+8)^{2/3)}