(\partial)/(\partial y)(x^3+y^3-3xy)

解答

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

3 y^{2} - 3 x

求解步骤

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

将

x

视为常数

使用微分加减法定则:

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

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

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

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

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

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

化简

= 3 y^{2} - 3 x

a re a f (x) = 4 - 3 x, g (x) = 2 x^{2} - x

\int_{- 3}^{- 2} x^{2} e^{x^{3} + 27} d x

x \to 2 lim (2 x - 9)

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

t an g e n t y = x (1.1)