(\partial)/(\partial x)(xe^{xy}-2y+1)

解答

\frac{\partial}{\partial x} (x e^{x y} - 2 y + 1)

e^{x y} + e^{x y} x y

求解步骤

\frac{\partial}{\partial x} (x e^{x y} - 2 y + 1)

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 e^{x y}) - \frac{\partial}{\partial x} (2 y) + \frac{\partial}{\partial x} (1)

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

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

\frac{\partial}{\partial x} (1) = 0

= e^{y x} + y e^{y x} x - 0 + 0

化简

= e^{x y} + e^{x y} x y

x \to 3 lim (cos (\frac{1}{x - 3}))

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

\frac{d}{d x} (0.005 x^{2} - ln (x^{16}) + 70)

d er i v a t i v e y = \frac{6 x ^{2} + 8 x + 2}{x}

d er i v a t i v e - 4 x^{\frac{1}{2}}