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

解

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

4 x^{3} - 4 y

解答ステップ

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

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

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

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

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

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

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

簡素化

= 4 x^{3} - 4 y