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

解

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

20 x^{4} - 24 x^{2} - 16 x y

解答ステップ

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

y

を定数として扱う

積の法則を適用:

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

f = 2 x^{3} - 4 y, g = 2 x^{2} - 4

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

\frac{\partial}{\partial x} (2 x^{3} - 4 y) = 6 x^{2}

\frac{\partial}{\partial x} (2 x^{2} - 4) = 4 x

= 6 x^{2} (2 x^{2} - 4) + 4 x (2 x^{3} - 4 y)

簡素化

6 x^{2} (2 x^{2} - 4) + 4 x (2 x^{3} - 4 y) : 20 x^{4} - 24 x^{2} - 16 x y

= 20 x^{4} - 24 x^{2} - 16 x y