What is the derivative of y= 3/(x^2)-8\sqrt[4]{x} ?

The derivative of y= 3/(x^2)-8\sqrt[4]{x} is (-6x^{3/4}-2x^3)/(x^{15/4)}

What is the first derivative of y= 3/(x^2)-8\sqrt[4]{x} ?

The first derivative of y= 3/(x^2)-8\sqrt[4]{x} is (-6x^{3/4}-2x^3)/(x^{15/4)}

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derivative of

y = \frac{3}{x ^{2}} - 8 4 x

\frac{- 6 x ^{\frac{3}{4}} - 2 x ^{3}}{x ^{\frac{15}{4}}}

Solution steps

\frac{d}{d x} (\frac{3}{x ^{2}} - 8 4 x)

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (\frac{3}{x ^{2}}) - \frac{d}{d x} (8 4 x)

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

\frac{d}{d x} (8 4 x) = \frac{2}{x ^{\frac{3}{4}}}

= - \frac{6}{x ^{3}} - \frac{2}{x ^{\frac{3}{4}}}

Simplify

- \frac{6}{x ^{3}} - \frac{2}{x ^{\frac{3}{4}}} : \frac{- 6 x ^{\frac{3}{4}} - 2 x ^{3}}{x ^{\frac{15}{4}}}

= \frac{- 6 x ^{\frac{3}{4}} - 2 x ^{3}}{x ^{\frac{15}{4}}}

\int \frac{x ^{2} + 5 x + 7}{x + 3} d x

\int y e^{4 y} d y

\frac{4 s e ^{- 3 s}}{( s ^{2} + 4 )}

\frac{d y}{d x}, y = 2 x - 5

n = 1 \sum \infty arctan (4 n)

What is the derivative of y= 3/(x^2)-8\sqrt[4]{x} ?
The derivative of y= 3/(x^2)-8\sqrt[4]{x} is (-6x^{3/4}-2x^3)/(x^{15/4)}
What is the first derivative of y= 3/(x^2)-8\sqrt[4]{x} ?
The first derivative of y= 3/(x^2)-8\sqrt[4]{x} is (-6x^{3/4}-2x^3)/(x^{15/4)}