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

解答

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

x (- x - 1)

求解步骤

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

将

x

视为常数

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

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

使用微分加减法定则:

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

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

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

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

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

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

= x (0 + 0 - x - 1)

化简

= x (- x - 1)

e x p an d (1 - t) s L + (1 - L) - \frac{( 1 - L ) ^{2}}{2} + xg

e x p an d x 3 (x^{2} + 1)^{2} + 3 (x^{2} + 1)^{2}

e x p an d (1 + t (x)^{2}) \frac{d ^{2}}{d t d x} (x) + x

e x p an d - \frac{3}{4} (e^{- x} - e^{3 x})

e x p an d \frac{V ( p + h ) ( V p - Vh ) ( h - p )}{p ^{2}}