What is the derivative of f(x)=\sqrt[4]{x^5}-3/(x^2) ?

The derivative of f(x)=\sqrt[4]{x^5}-3/(x^2) is (5x^4)/(4(x^5)^{3/4)}+6/(x^3)

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

The first derivative of f(x)=\sqrt[4]{x^5}-3/(x^2) is (5x^4)/(4(x^5)^{3/4)}+6/(x^3)

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derivative of

\frac{5 x ^{4}}{4 ( x ^{5} ) ^{\frac{3}{4}}} + \frac{6}{x ^{3}}

Solution steps

Apply the Sum/Difference Rule:

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

\frac{d}{d x} (\frac{3}{x ^{2}}) = - \frac{6}{x ^{3}}

= \frac{5 x ^{4}}{4 ( x ^{5} ) ^{\frac{3}{4}}} - (- \frac{6}{x ^{3}})

Simplify

= \frac{5 x ^{4}}{4 ( x ^{5} ) ^{\frac{3}{4}}} + \frac{6}{x ^{3}}

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What is the derivative of f(x)=\sqrt[4]{x^5}-3/(x^2) ?
The derivative of f(x)=\sqrt[4]{x^5}-3/(x^2) is (5x^4)/(4(x^5)^{3/4)}+6/(x^3)
What is the first derivative of f(x)=\sqrt[4]{x^5}-3/(x^2) ?
The first derivative of f(x)=\sqrt[4]{x^5}-3/(x^2) is (5x^4)/(4(x^5)^{3/4)}+6/(x^3)