What is the derivative of 1/3 x^3-5x-4/x ?

The derivative of 1/3 x^3-5x-4/x is x^2-5+4/(x^2)

What is the first derivative of 1/3 x^3-5x-4/x ?

The first derivative of 1/3 x^3-5x-4/x is x^2-5+4/(x^2)

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\frac{d}{d x} (\frac{1}{3} x^{3} - 5 x - \frac{4}{x})

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

Solution steps

\frac{d}{d x} (\frac{1}{3} x^{3} - 5 x - \frac{4}{x})

Simplify

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

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

Apply the Sum/Difference Rule:

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

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

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

\frac{d}{d x} (5 x) = 5

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

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

Simplify

= x^{2} - 5 + \frac{4}{x ^{2}}

\int x \cdot ln^{5} (x) d x

y^{^{''}} - 4 y^{^{'}} + 4 y = e^{x} + x e^{2 x}

f (x) = cot (x^{2})

\int_{- 1}^{0} x^{3} - x^{2} - 2 x d x

\int x^{4} cos (2 x) d x

What is the derivative of 1/3 x^3-5x-4/x ?
The derivative of 1/3 x^3-5x-4/x is x^2-5+4/(x^2)
What is the first derivative of 1/3 x^3-5x-4/x ?
The first derivative of 1/3 x^3-5x-4/x is x^2-5+4/(x^2)