derivative of x^3+2x^2-x+4

解答

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

3 x^{2} + 4 x - 1

求解步骤

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

使用微分加减法定则:

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

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

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

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

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

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

= 3 x^{2} + 4 x - 1 + 0

化简

= 3 x^{2} + 4 x - 1

\int \frac{1}{x ^{2} - x ^{3}} d x

\frac{\partial}{\partial x} (ln (2 x - y))

\int_{0}^{9} 9 + 3 x d x

d er i v a t i v e 3 t^{2}

y^{^{''}} - 4 y^{^{'}} + 53 y = 5 e^{2 t} cos (7 t)