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

The derivative of \sqrt[3]{(1+x^3/(1-x^3)}) is (2x^2)/((x^3+1)^{2/3)(-x^3+1)^{4/3}}

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

The first derivative of \sqrt[3]{(1+x^3/(1-x^3)}) is (2x^2)/((x^3+1)^{2/3)(-x^3+1)^{4/3}}

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

Solution steps

Apply the chain rule:

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

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

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

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

Simplify

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

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

What is the derivative of \sqrt[3]{(1+x^3/(1-x^3)}) ?
The derivative of \sqrt[3]{(1+x^3/(1-x^3)}) is (2x^2)/((x^3+1)^{2/3)(-x^3+1)^{4/3}}
What is the first derivative of \sqrt[3]{(1+x^3/(1-x^3)}) ?
The first derivative of \sqrt[3]{(1+x^3/(1-x^3)}) is (2x^2)/((x^3+1)^{2/3)(-x^3+1)^{4/3}}