What is the derivative of \sqrt[4]{x}+1/(x^4) ?

The derivative of \sqrt[4]{x}+1/(x^4) is (x^5-16x^{3/4})/(4x^{23/4)}

What is the first derivative of \sqrt[4]{x}+1/(x^4) ?

The first derivative of \sqrt[4]{x}+1/(x^4) is (x^5-16x^{3/4})/(4x^{23/4)}

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derivative of

\frac{x ^{5} - 16 x ^{\frac{3}{4}}}{4 x ^{\frac{23}{4}}}

Solution steps

Apply the Sum/Difference Rule:

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

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

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

Simplify

\frac{1}{4 x ^{\frac{3}{4}}} - \frac{4}{x ^{5}} : \frac{x ^{5} - 16 x ^{\frac{3}{4}}}{4 x ^{\frac{23}{4}}}

= \frac{x ^{5} - 16 x ^{\frac{3}{4}}}{4 x ^{\frac{23}{4}}}

What is the derivative of \sqrt[4]{x}+1/(x^4) ?
The derivative of \sqrt[4]{x}+1/(x^4) is (x^5-16x^{3/4})/(4x^{23/4)}
What is the first derivative of \sqrt[4]{x}+1/(x^4) ?
The first derivative of \sqrt[4]{x}+1/(x^4) is (x^5-16x^{3/4})/(4x^{23/4)}