(\partial)/(\partial x)(-6y^3+10x^2-4xy)

解

\frac{\partial}{\partial x} (- 6 y^{3} + 10 x^{2} - 4 x y)

20 x - 4 y

解答ステップ

\frac{\partial}{\partial x} (- 6 y^{3} + 10 x^{2} - 4 x y)

y

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 x} (6 y^{3}) + \frac{\partial}{\partial x} (10 x^{2}) - \frac{\partial}{\partial x} (4 x y)

\frac{\partial}{\partial x} (6 y^{3}) = 0

\frac{\partial}{\partial x} (10 x^{2}) = 20 x

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

= - 0 + 20 x - 4 y

簡素化

= 20 x - 4 y