(\partial)/(\partial x)(-x^3+4xy-2y^2+1)

解

\frac{\partial}{\partial x} (- x^{3} + 4 x y - 2 y^{2} + 1)

4 y - 3 x^{2}

解答ステップ

\frac{\partial}{\partial x} (- x^{3} + 4 x y - 2 y^{2} + 1)

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} (x^{3}) + \frac{\partial}{\partial x} (4 x y) - \frac{\partial}{\partial x} (2 y^{2}) + \frac{\partial}{\partial x} (1)

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

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

\frac{\partial}{\partial x} (2 y^{2}) = 0

\frac{\partial}{\partial x} (1) = 0

= - 3 x^{2} + 4 y - 0 + 0

簡素化

= 4 y - 3 x^{2}