(\partial)/(\partial y)(xy+yz+zx)

解答

\frac{\partial}{\partial y} (x y + yz + z x)

x + z

求解步骤

\frac{\partial}{\partial y} (x y + yz + z x)

将

x, z

视为常数

使用微分加减法定则:

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

= \frac{\partial}{\partial y} (x y) + \frac{\partial}{\partial y} (yz) + \frac{\partial}{\partial y} (z x)

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

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

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

= x + z + 0

化简

= x + z

x \to 5 lim (\frac{f ( x )}{g ( x )})

\int_{0}^{2} 6 - x^{2} d x

\frac{\partial}{\partial y} (2 x + 3 y)

a re a y = 4 x, y = \frac{4 x}{3}

\frac{\partial}{\partial x} (cos (6 x))