(\partial)/(\partial y)(z^3-xy+yz+y^3-2)

解答

\frac{\partial}{\partial y} (z^{3} - x y + yz + y^{3} - 2)

z + 3 y^{2} - x

求解步骤

\frac{\partial}{\partial y} (z^{3} - x y + yz + y^{3} - 2)

将

z, x

视为常数

使用微分加减法定则:

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

= \frac{\partial}{\partial y} (z^{3}) - \frac{\partial}{\partial y} (x y) + \frac{\partial}{\partial y} (yz) + \frac{\partial}{\partial y} (y^{3}) - \frac{\partial}{\partial y} (2)

\frac{\partial}{\partial y} (z^{3}) = 0

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

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

\frac{\partial}{\partial y} (y^{3}) = 3 y^{2}

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

= 0 - x + z + 3 y^{2} - 0

化简

= z + 3 y^{2} - x

\int \frac{- 11 x}{( x ^{2} + 5 ) ^{2}} d x

\int \frac{1}{s ^{2} + 3 s + 6} d s

\frac{d}{d x} (tan^{- 1} (x))

a re a 2 x, x^{2}, 0, 2

\int \frac{1}{2} e^{- 2 x} d x