What is the derivative of \sqrt[5]{x}-3/x ?

The derivative of \sqrt[5]{x}-3/x is (x^2+15x^{4/5})/(5x^{14/5)}

What is the first derivative of \sqrt[5]{x}-3/x ?

The first derivative of \sqrt[5]{x}-3/x is (x^2+15x^{4/5})/(5x^{14/5)}

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\frac{x ^{2} + 15 x ^{\frac{4}{5}}}{5 x ^{\frac{14}{5}}}

Solution steps

Apply the Sum/Difference Rule:

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

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

= \frac{1}{5 x ^{\frac{4}{5}}} - (- \frac{3}{x ^{2}})

Simplify

\frac{1}{5 x ^{\frac{4}{5}}} - (- \frac{3}{x ^{2}}) : \frac{x ^{2} + 15 x ^{\frac{4}{5}}}{5 x ^{\frac{14}{5}}}

= \frac{x ^{2} + 15 x ^{\frac{4}{5}}}{5 x ^{\frac{14}{5}}}

What is the derivative of \sqrt[5]{x}-3/x ?
The derivative of \sqrt[5]{x}-3/x is (x^2+15x^{4/5})/(5x^{14/5)}
What is the first derivative of \sqrt[5]{x}-3/x ?
The first derivative of \sqrt[5]{x}-3/x is (x^2+15x^{4/5})/(5x^{14/5)}