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

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

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

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

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

Solution steps

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

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