(\partial)/(\partial x)(xe^y+y+1)

解答

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

e^{y}

求解步骤

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

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

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

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

= e^{y} + 0 + 0

化简

= e^{y}

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

\int_{0}^{0.02} 200 x d x

in v erse l a pl a ce \frac{3 s + 1}{s ( s + 2 )}

t an g e n t \frac{1}{2 x + 2}, at x = 5

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