(\partial)/(\partial x)(3x^2-y^3+x-1)

解

\frac{\partial}{\partial x} (3 x^{2} - y^{3} + x - 1)

6 x + 1

解答ステップ

\frac{\partial}{\partial x} (3 x^{2} - y^{3} + 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} (3 x^{2}) - \frac{\partial}{\partial x} (y^{3}) + \frac{\partial}{\partial x} (x) - \frac{\partial}{\partial x} (1)

\frac{\partial}{\partial x} (3 x^{2}) = 6 x

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

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

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

= 6 x - 0 + 1 - 0

簡素化

= 6 x + 1