What is the derivative of y=-2x^{-4}+4/x-(2x^2)/3 ?

The derivative of y=-2x^{-4}+4/x-(2x^2)/3 is 8/(x^5)-4/(x^2)-(4x)/3

What is the first derivative of y=-2x^{-4}+4/x-(2x^2)/3 ?

The first derivative of y=-2x^{-4}+4/x-(2x^2)/3 is 8/(x^5)-4/(x^2)-(4x)/3

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derivative of

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

\frac{8}{x ^{5}} - \frac{4}{x ^{2}} - \frac{4 x}{3}

Solution steps

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

Apply the Sum/Difference Rule:

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

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

\frac{d}{d x} (2 x^{- 4}) = - \frac{8}{x ^{5}}

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

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

= - (- \frac{8}{x ^{5}}) - \frac{4}{x ^{2}} - \frac{4 x}{3}

Simplify

= \frac{8}{x ^{5}} - \frac{4}{x ^{2}} - \frac{4 x}{3}

\frac{d}{d x} (x^{2} + 4 x + 4)

\int x 3^{x} d x

\frac{d}{d x} (\frac{csc ( x )}{cot ( x )})

\int \frac{2 x ^{4} - 8 x ^{3} - 6 x ^{2} + 4}{x ^{3}} d x

y^{^{''}} - 4 y^{^{'}} + 4 y = t^{- 5} e^{2 t}

What is the derivative of y=-2x^{-4}+4/x-(2x^2)/3 ?
The derivative of y=-2x^{-4}+4/x-(2x^2)/3 is 8/(x^5)-4/(x^2)-(4x)/3
What is the first derivative of y=-2x^{-4}+4/x-(2x^2)/3 ?
The first derivative of y=-2x^{-4}+4/x-(2x^2)/3 is 8/(x^5)-4/(x^2)-(4x)/3