(\partial)/(\partial y)(e^{xyz^2}yz^2)

解

\frac{\partial}{\partial y} (e^{x y z^{2}} y z^{2})

z^{2} (e^{x y z^{2}} x y z^{2} + e^{x y z^{2}})

解答ステップ

\frac{\partial}{\partial y} (e^{x y z^{2}} y z^{2})

x, z

を定数として扱う

定数を除去:

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

= z^{2} \frac{\partial}{\partial y} (e^{x y z^{2}} y)

積の法則を適用:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = e^{x y z^{2}}, g = y

= z^{2} (\frac{\partial}{\partial y} (e^{x y z^{2}}) y + \frac{\partial}{\partial y} (y) e^{x y z^{2}})

\frac{\partial}{\partial y} (e^{x y z^{2}}) = e^{x y z^{2}} x z^{2}

\frac{\partial}{\partial y} (y) = 1

= z^{2} (e^{x y z^{2}} x z^{2} y + 1 \cdot e^{x y z^{2}})

簡素化

= z^{2} (e^{x y z^{2}} x y z^{2} + e^{x y z^{2}})