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

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

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

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

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

Solution steps

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d y}{d x} + \frac{3}{x} y = 16 y^{3}

\int_{4}^{7} \frac{x}{x ^{2} + 6 x + 13} d x

f (x) = 2 x^{2} + 2 x, at x = - 3

\int x e - x d x

\int r^{2} sin (θ) d r

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