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{d}{d x} (3 x^{3} - 3 x + c)

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

Solution steps

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

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}}}

\frac{d y}{d x} - 7 = 3 y - 7 x + 9

3 x - 1

\int + 5 arctan (\frac{1}{x}) d x

\int \frac{sec ^{2} ( x )}{tan ^{6} ( x )} d x

n = 1 \sum \infty \frac{1}{n ^{1.01}}

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)}