(\partial)/(\partial y)(xy^3+4ye^x-2y^2)

解

\frac{\partial}{\partial y} (x y^{3} + 4 y e^{x} - 2 y^{2})

3 x y^{2} + 4 e^{x} - 4 y

解答ステップ

\frac{\partial}{\partial y} (x y^{3} + 4 y e^{x} - 2 y^{2})

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} (x y^{3}) + \frac{\partial}{\partial y} (4 y e^{x}) - \frac{\partial}{\partial y} (2 y^{2})

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

\frac{\partial}{\partial y} (4 y e^{x}) = 4 e^{x}

\frac{\partial}{\partial y} (2 y^{2}) = 4 y

= 3 x y^{2} + 4 e^{x} - 4 y