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

解

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

3 x^{2} y^{2} + 3 y - 8 x y

解答ステップ

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

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

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

\frac{\partial}{\partial x} (4 y x^{2}) = 8 y x

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

\frac{\partial}{\partial x} (4) = 0

= 3 y^{2} x^{2} - 8 y x + 3 y - 0

簡素化

= 3 x^{2} y^{2} + 3 y - 8 x y