(\partial)/(\partial x)(y^2-2xy-2x^2)

解答

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

- 2 y - 4 x

求解步骤

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

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

\frac{\partial}{\partial x} (y^{2}) = 0

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

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

= 0 - 2 y - 4 x

化简

= - 2 y - 4 x

d er i v a t i v e f (x) = 5 x + 7

(\frac{x ^{2} + 1}{x ^{2} - 1})^{^{'}}

x \to 2.99 lim (1 + x)

\int \frac{1}{1 + 1 + x} d x

d er i v a t i v e 3 x^{2} - 6