What is the derivative of 6\sqrt[4]{x}-5/(x^2) ?

The derivative of 6\sqrt[4]{x}-5/(x^2) is (3x^3+20x^{3/4})/(2x^{15/4)}

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

The first derivative of 6\sqrt[4]{x}-5/(x^2) is (3x^3+20x^{3/4})/(2x^{15/4)}

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derivative of

6 4 x - \frac{5}{x ^{2}}

\frac{3 x ^{3} + 20 x ^{\frac{3}{4}}}{2 x ^{\frac{15}{4}}}

Solution steps

\frac{d}{d x} (6 4 x - \frac{5}{x ^{2}})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (6 4 x) - \frac{d}{d x} (\frac{5}{x ^{2}})

\frac{d}{d x} (6 4 x) = \frac{3}{2 x ^{\frac{3}{4}}}

\frac{d}{d x} (\frac{5}{x ^{2}}) = - \frac{10}{x ^{3}}

= \frac{3}{2 x ^{\frac{3}{4}}} - (- \frac{10}{x ^{3}})

Simplify

\frac{3}{2 x ^{\frac{3}{4}}} - (- \frac{10}{x ^{3}}) : \frac{3 x ^{3} + 20 x ^{\frac{3}{4}}}{2 x ^{\frac{15}{4}}}

= \frac{3 x ^{3} + 20 x ^{\frac{3}{4}}}{2 x ^{\frac{15}{4}}}

\int \frac{x + 36}{x} d x

\int csc (θ) cot (θ) d θ

d er i v a t i v e f (x) = 6 x

\int 3 x^{2} sin (x^{3}) d x

\frac{\partial}{\partial x} (\frac{2}{3 x})

What is the derivative of 6\sqrt[4]{x}-5/(x^2) ?
The derivative of 6\sqrt[4]{x}-5/(x^2) is (3x^3+20x^{3/4})/(2x^{15/4)}
What is the first derivative of 6\sqrt[4]{x}-5/(x^2) ?
The first derivative of 6\sqrt[4]{x}-5/(x^2) is (3x^3+20x^{3/4})/(2x^{15/4)}