(\partial)/(\partial x)(40xy*e^{5x^2-2y})

解

\frac{\partial}{\partial x} (40 x y \cdot e^{5 x^{2} - 2 y})

40 y (e^{5 x^{2} - 2 y} + 10 e^{5 x^{2} - 2 y} x^{2})

解答ステップ

\frac{\partial}{\partial x} (40 x y e^{5 x^{2} - 2 y})

y

を定数として扱う

定数を除去:

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

= 40 y \frac{\partial}{\partial x} (x e^{5 x^{2} - 2 y})

積の法則を適用:

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

f = x, g = e^{5 x^{2} - 2 y}

= 40 y (\frac{\partial}{\partial x} (x) e^{5 x^{2} - 2 y} + \frac{\partial}{\partial x} (e^{5 x^{2} - 2 y}) x)

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

\frac{\partial}{\partial x} (e^{5 x^{2} - 2 y}) = e^{5 x^{2} - 2 y} \cdot 10 x

= 40 y (1 \cdot e^{5 x^{2} - 2 y} + e^{5 x^{2} - 2 y} \cdot 10 xx)

簡素化

40 y (1 \cdot e^{5 x^{2} - 2 y} + e^{5 x^{2} - 2 y} \cdot 10 xx) : 40 y (e^{5 x^{2} - 2 y} + 10 e^{5 x^{2} - 2 y} x^{2})

= 40 y (e^{5 x^{2} - 2 y} + 10 e^{5 x^{2} - 2 y} x^{2})