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

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

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

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

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

Solution steps

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

(10 - 6 y + e^{- 3 x}) d x - 2 d y = 0

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

\int arcsin (\frac{x}{x + 1}) d x

\frac{\partial}{\partial x} (e^{- \frac{1}{2} x y})

\int - 2 x + 6 d x

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