d/(dy)(x^2-xy+2)

解答

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

- x

求解步骤

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

将

x

视为常数

使用微分加减法定则:

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

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

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

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

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

= 0 - x + 0

化简

= - x

d er i v a t i v e y = \frac{5 + x f ( x )}{x}

\int 9 x^{4} d x

y^{^{'}} + 2 x y = 2 x e^{- x^{2}}

\frac{d y}{d x} = \frac{y ^{2} x}{x ^{2} + 1}

y^{^{''}} - 8 y^{^{'}} + 16 y = \frac{12 e ^{4 t}}{( t ^{2} + 1 )}