(\partial)/(\partial y)(zln(y)+e^2)

解答

\frac{\partial}{\partial y} (z ln (y) + e^{2})

\frac{z}{y}

求解步骤

\frac{\partial}{\partial y} (z ln (y) + e^{2})

将

z

视为常数

使用微分加减法定则:

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

= \frac{\partial}{\partial y} (z ln (y)) + \frac{\partial}{\partial y} (e^{2})

\frac{\partial}{\partial y} (z ln (y)) = \frac{z}{y}

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

= \frac{z}{y} + 0

化简

= \frac{z}{y}

\frac{d}{d x} (- x e^{\frac{y}{x}})

\int_{0}^{\frac{π}{2}} cot (x) d x

x \to 0 lim ((1 + 8 x)^{\frac{6}{x}})

s im pl i f y \frac{4 - 3 x - x ^{2}}{x ^{2} - 1}

\frac{d x}{d y} = 3 x