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

解答

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

3 x^{2} y^{2} + 3 y - 8 x y

求解步骤

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

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

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

\frac{\partial}{\partial x} (4 y x^{2}) = 8 y x

\frac{\partial}{\partial x} (3 y x) = 3 y

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

= 3 y^{2} x^{2} - 8 y x + 3 y - 0

化简

= 3 x^{2} y^{2} + 3 y - 8 x y

t an g e n t y = x^{2} + 5 x + 6, at x = 7

\int 144 - 9 x^{2} d x

x \to - 5 + lim (\frac{3}{x ^{2} - 25})

\frac{d}{d x} (x^{s i n (3 x)})

\frac{\partial}{\partial u} (cos (u + v))