What is the derivative of \sqrt[3]{x}-9/x ?

The derivative of \sqrt[3]{x}-9/x is (x^2+27x^{2/3})/(3x^{8/3)}

What is the first derivative of \sqrt[3]{x}-9/x ?

The first derivative of \sqrt[3]{x}-9/x is (x^2+27x^{2/3})/(3x^{8/3)}

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\frac{x ^{2} + 27 x ^{\frac{2}{3}}}{3 x ^{\frac{8}{3}}}

Solution steps

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

\frac{d}{d x} (\frac{9}{x}) = - \frac{9}{x ^{2}}

= \frac{1}{3 x ^{\frac{2}{3}}} - (- \frac{9}{x ^{2}})

Simplify

\frac{1}{3 x ^{\frac{2}{3}}} - (- \frac{9}{x ^{2}}) : \frac{x ^{2} + 27 x ^{\frac{2}{3}}}{3 x ^{\frac{8}{3}}}

= \frac{x ^{2} + 27 x ^{\frac{2}{3}}}{3 x ^{\frac{8}{3}}}

n = 0 \sum \infty 2 + (0.1)^{n}

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What is the derivative of \sqrt[3]{x}-9/x ?
The derivative of \sqrt[3]{x}-9/x is (x^2+27x^{2/3})/(3x^{8/3)}
What is the first derivative of \sqrt[3]{x}-9/x ?
The first derivative of \sqrt[3]{x}-9/x is (x^2+27x^{2/3})/(3x^{8/3)}