What is the derivative of \sqrt[3]{x^3-3x+c} ?

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

What is the first derivative of \sqrt[3]{x^3-3x+c} ?

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

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

Solution steps

Apply the chain rule:

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

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

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

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

Simplify

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

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

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