What is the derivative of \sqrt[7]{x}-4/x ?

The derivative of \sqrt[7]{x}-4/x is (x^2+28x^{6/7})/(7x^{20/7)}

What is the first derivative of \sqrt[7]{x}-4/x ?

The first derivative of \sqrt[7]{x}-4/x is (x^2+28x^{6/7})/(7x^{20/7)}

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\frac{x ^{2} + 28 x ^{\frac{6}{7}}}{7 x ^{\frac{20}{7}}}

Solution steps

Apply the Sum/Difference Rule:

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

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

= \frac{1}{7 x ^{\frac{6}{7}}} - (- \frac{4}{x ^{2}})

Simplify

\frac{1}{7 x ^{\frac{6}{7}}} - (- \frac{4}{x ^{2}}) : \frac{x ^{2} + 28 x ^{\frac{6}{7}}}{7 x ^{\frac{20}{7}}}

= \frac{x ^{2} + 28 x ^{\frac{6}{7}}}{7 x ^{\frac{20}{7}}}

What is the derivative of \sqrt[7]{x}-4/x ?
The derivative of \sqrt[7]{x}-4/x is (x^2+28x^{6/7})/(7x^{20/7)}
What is the first derivative of \sqrt[7]{x}-4/x ?
The first derivative of \sqrt[7]{x}-4/x is (x^2+28x^{6/7})/(7x^{20/7)}