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

The derivative of (-3x^5-x^4-x^3+4x^2/(x^2)) is -9x^2-2x-1

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

The first derivative of (-3x^5-x^4-x^3+4x^2/(x^2)) is -9x^2-2x-1

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\frac{d}{d x} (\frac{- 3 x ^{5} - x ^{4} - x ^{3} + 4 x ^{2}}{x ^{2}})

- 9 x^{2} - 2 x - 1

Solution steps

\frac{d}{d x} (\frac{- 3 x ^{5} - x ^{4} - x ^{3} + 4 x ^{2}}{x ^{2}})

Simplify

\frac{- 3 x ^{5} - x ^{4} - x ^{3} + 4 x ^{2}}{x ^{2}} : - 3 x^{3} - x^{2} - x + 4

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

Apply the Sum/Difference Rule:

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

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

\frac{d}{d x} (3 x^{3}) = 9 x^{2}

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

\frac{d x}{d x} = 1

\frac{d}{d x} (4) = 0

= - 9 x^{2} - 2 x - 1 + 0

Simplify

= - 9 x^{2} - 2 x - 1

y^{^{''}} + 4 y^{^{'}} + 4 y = 0, y (0) = - 1, y^{^{'}} (0) = 0

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

\int_{0}^{4} \frac{1}{x ^{2} + x - 6} d x

\int_{1}^{4} (4 x^{2} + x) d x

\int \frac{1}{1 - 5 t} d t

What is the derivative of (-3x^5-x^4-x^3+4x^2/(x^2)) ?
The derivative of (-3x^5-x^4-x^3+4x^2/(x^2)) is -9x^2-2x-1
What is the first derivative of (-3x^5-x^4-x^3+4x^2/(x^2)) ?
The first derivative of (-3x^5-x^4-x^3+4x^2/(x^2)) is -9x^2-2x-1