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

解

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

z^{2}

解答ステップ

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

z, t, x

を定数として扱う

定数を除去:

(a \cdot f)^{'} = a \cdot f^{'}

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

和/差の法則を適用:

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

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

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

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

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

= z (0 + z - 0)

簡素化

z (0 + z - 0) : z^{2}

= z^{2}