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

解

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

z + 3 y^{2} - x

解答ステップ

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

z, x

を定数として扱う

和/差の法則を適用:

(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

簡素化

= z + 3 y^{2} - x