What is the derivative of \sqrt[8]{x}-9/x ?

The derivative of \sqrt[8]{x}-9/x is (x^2+72x^{7/8})/(8x^{23/8)}

What is the first derivative of \sqrt[8]{x}-9/x ?

The first derivative of \sqrt[8]{x}-9/x is (x^2+72x^{7/8})/(8x^{23/8)}

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\frac{x ^{2} + 72 x ^{\frac{7}{8}}}{8 x ^{\frac{23}{8}}}

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}{8 x ^{\frac{7}{8}}} - (- \frac{9}{x ^{2}})

Simplify

\frac{1}{8 x ^{\frac{7}{8}}} - (- \frac{9}{x ^{2}}) : \frac{x ^{2} + 72 x ^{\frac{7}{8}}}{8 x ^{\frac{23}{8}}}

= \frac{x ^{2} + 72 x ^{\frac{7}{8}}}{8 x ^{\frac{23}{8}}}

What is the derivative of \sqrt[8]{x}-9/x ?
The derivative of \sqrt[8]{x}-9/x is (x^2+72x^{7/8})/(8x^{23/8)}
What is the first derivative of \sqrt[8]{x}-9/x ?
The first derivative of \sqrt[8]{x}-9/x is (x^2+72x^{7/8})/(8x^{23/8)}