What is the derivative of \sqrt[3]{x^3-1/(x^3}) ?

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

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

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

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Popular >

\frac{d}{d x} (3 x^{3} - \frac{1}{x ^{3}})

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

Solution steps

\frac{d}{d x} (3 x^{3} - \frac{1}{x ^{3}})

Apply the chain rule:

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

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

\frac{d}{d x} (x^{3} - \frac{1}{x ^{3}}) = 3 x^{2} + \frac{3}{x ^{4}}

= \frac{1}{3 ( x ^{3} - \frac{1}{x ^{3}} ) ^{\frac{2}{3}}} (3 x^{2} + \frac{3}{x ^{4}})

Simplify

\frac{1}{3 ( x ^{3} - \frac{1}{x ^{3}} ) ^{\frac{2}{3}}} (3 x^{2} + \frac{3}{x ^{4}}) : \frac{x ^{6} + 1}{x ^{2} ( x ^{6} - 1 ) ^{\frac{2}{3}}}

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

e^{- 4 t} \cdot cos (3 t)

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

\int cos^{2} (8 x) d x

\int θ^{2} d θ

n = 1 \sum \infty (- 1)^{n} n^{2}

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