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

解

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

2 x + 1

解答ステップ

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

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

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

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

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

= 2 x - 0 + 1 + 0

簡素化

= 2 x + 1