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

解

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

- 4 x^{3} + 4 x

解答ステップ

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

\frac{\partial}{\partial x} (3) = 0

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

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

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

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

簡素化

= - 4 x^{3} + 4 x