(\partial)/(\partial x)(4y^3+2x^2y+1)

解答

\frac{\partial}{\partial x} (4 y^{3} + 2 x^{2} y + 1)

4 y x

求解步骤

\frac{\partial}{\partial x} (4 y^{3} + 2 x^{2} y + 1)

将

y

视为常数

使用微分加减法定则:

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

= \frac{\partial}{\partial x} (4 y^{3}) + \frac{\partial}{\partial x} (2 x^{2} y) + \frac{\partial}{\partial x} (1)

\frac{\partial}{\partial x} (4 y^{3}) = 0

\frac{\partial}{\partial x} (2 x^{2} y) = 4 y x

\frac{\partial}{\partial x} (1) = 0

= 0 + 4 y x + 0

化简

= 4 y x

\frac{d y}{d x} x + 7 y = e^{x^{7}}

\int (y e^{x^{2} + y^{2}} + 3 y^{2}) d y

\frac{d}{d x} (\frac{x ^{5}}{6} + \frac{1}{10 x ^{3}})

d er i v a t i v e f (x) = \frac{2 - 3 x - x ^{2}}{x ^{2} - 4}

\int \frac{4 x}{x ^{2} - 1} d x