What is the derivative of \sqrt[5]{x^3}-7/x ?

The derivative of \sqrt[5]{x^3}-7/x is (3x^2)/(5(x^3)^{4/5)}+7/(x^2)

What is the first derivative of \sqrt[5]{x^3}-7/x ?

The first derivative of \sqrt[5]{x^3}-7/x is (3x^2)/(5(x^3)^{4/5)}+7/(x^2)

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derivative of

\frac{3 x ^{2}}{5 ( x ^{3} ) ^{\frac{4}{5}}} + \frac{7}{x ^{2}}

Solution steps

Apply the Sum/Difference Rule:

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

\frac{d}{d x} (\frac{7}{x}) = - \frac{7}{x ^{2}}

= \frac{3 x ^{2}}{5 ( x ^{3} ) ^{\frac{4}{5}}} - (- \frac{7}{x ^{2}})

Simplify

= \frac{3 x ^{2}}{5 ( x ^{3} ) ^{\frac{4}{5}}} + \frac{7}{x ^{2}}

What is the derivative of \sqrt[5]{x^3}-7/x ?
The derivative of \sqrt[5]{x^3}-7/x is (3x^2)/(5(x^3)^{4/5)}+7/(x^2)
What is the first derivative of \sqrt[5]{x^3}-7/x ?
The first derivative of \sqrt[5]{x^3}-7/x is (3x^2)/(5(x^3)^{4/5)}+7/(x^2)