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

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

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

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

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derivative of

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

Solution steps

Apply the chain rule:

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

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

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

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

Simplify

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

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

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