(\partial)/(\partial y)(x^3-3x+y-e^y)

解

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

- e^{y} + 1

解答ステップ

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

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

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

\frac{\partial}{\partial y} (3 x) = 0

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

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

= 0 - 0 + 1 - e^{y}

簡素化

= - e^{y} + 1