(\partial)/(\partial x)((x-7)^2+(y-1)^2+29)

解

\frac{\partial}{\partial x} ((x - 7)^{2} + (y - 1)^{2} + 29)

2 x - 14

解答ステップ

\frac{\partial}{\partial x} ((x - 7)^{2} + (y - 1)^{2} + 29)

簡素化

(x - 7)^{2} + (y - 1)^{2} + 29 : x^{2} - 14 x + y^{2} + 79 - 2 y

= \frac{\partial}{\partial x} (x^{2} - 14 x + y^{2} + 79 - 2 y)

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} (14 x) + \frac{\partial}{\partial x} (y^{2}) + \frac{\partial}{\partial x} (79) - \frac{\partial}{\partial x} (2 y)

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

\frac{\partial}{\partial x} (14 x) = 14

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

\frac{\partial}{\partial x} (79) = 0

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

= 2 x - 14 + 0 + 0 - 0

簡素化

= 2 x - 14