(\partial)/(\partial x)(x^2-2xy+6y^2+x-5y)

解答

\frac{\partial}{\partial x} (x^{2} - 2 x y + 6 y^{2} + x - 5 y)

2 x + 1 - 2 y

求解步骤

\frac{\partial}{\partial x} (x^{2} - 2 x y + 6 y^{2} + x - 5 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^{2}) - \frac{\partial}{\partial x} (2 x y) + \frac{\partial}{\partial x} (6 y^{2}) + \frac{\partial}{\partial x} (x) - \frac{\partial}{\partial x} (5 y)

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

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

\frac{\partial}{\partial x} (6 y^{2}) = 0

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

\frac{\partial}{\partial x} (5 y) = 0

= 2 x - 2 y + 0 + 1 - 0

化简

= 2 x + 1 - 2 y

\int y^{5} d y

x \to 0 lim (\frac{e ^{x} + 1}{x})

t an g e n t 3 x^{2} - 5 x

\int \frac{e ^{4 x}}{e ^{4 x} + 3} d x

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