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

解答

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

3 x^{2} - 3 y

求解步骤

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

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

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

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

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

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

化简

= 3 x^{2} - 3 y

t an g e n t f (x) = 7 e^{x} cos (x), at x = 0

x \to 0 lim ((\frac{x - 4}{x + 1})^{x - 2})

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

y^{^{'}} + 3 y = e^{- 4 t}, y (0) = 5

\frac{d}{d x} (\frac{9}{e ^{x} - e ^{- x}})