What is the derivative of \sqrt[5]{(x^3-4x^{-3}}) ?

The derivative of \sqrt[5]{(x^3-4x^{-3}}) is -(3(3x^2-4))/(5(x^3-4x)^{8/5)}

What is the first derivative of \sqrt[5]{(x^3-4x^{-3}}) ?

The first derivative of \sqrt[5]{(x^3-4x^{-3}}) is -(3(3x^2-4))/(5(x^3-4x)^{8/5)}

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- \frac{3 ( 3 x ^{2} - 4 )}{5 ( x ^{3} - 4 x ) ^{\frac{8}{5}}}

Solution steps

Apply the chain rule:

\frac{1}{5 ( ( x ^{3} - 4 x ) ^{- 3} ) ^{\frac{4}{5}}} \frac{d}{d x} ((x^{3} - 4 x)^{- 3})

= \frac{1}{5 ( ( x ^{3} - 4 x ) ^{- 3} ) ^{\frac{4}{5}}} \frac{d}{d x} ((x^{3} - 4 x)^{- 3})

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

= \frac{1}{5 ( ( x ^{3} - 4 x ) ^{- 3} ) ^{\frac{4}{5}}} (- \frac{3 ( 3 x ^{2} - 4 )}{( x ^{3} - 4 x ) ^{4}})

Simplify

\frac{1}{5 ( ( x ^{3} - 4 x ) ^{- 3} ) ^{\frac{4}{5}}} (- \frac{3 ( 3 x ^{2} - 4 )}{( x ^{3} - 4 x ) ^{4}}) : - \frac{3 ( 3 x ^{2} - 4 )}{5 ( x ^{3} - 4 x ) ^{\frac{8}{5}}}

= - \frac{3 ( 3 x ^{2} - 4 )}{5 ( x ^{3} - 4 x ) ^{\frac{8}{5}}}

What is the derivative of \sqrt[5]{(x^3-4x^{-3}}) ?
The derivative of \sqrt[5]{(x^3-4x^{-3}}) is -(3(3x^2-4))/(5(x^3-4x)^{8/5)}
What is the first derivative of \sqrt[5]{(x^3-4x^{-3}}) ?
The first derivative of \sqrt[5]{(x^3-4x^{-3}}) is -(3(3x^2-4))/(5(x^3-4x)^{8/5)}