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

解

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

8 x y^{3} + 1

解答ステップ

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

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

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

\frac{\partial}{\partial x} (3 y^{2}) = 0

\frac{\partial}{\partial x} (4 x^{2} y^{3}) = 8 y^{3} x

= 1 - 0 + 8 y^{3} x

簡素化

= 8 x y^{3} + 1