(\partial)/(\partial x)(x^2+y^2+(21-x-y)^2)

解

\frac{\partial}{\partial x} (x^{2} + y^{2} + (21 - x - y)^{2})

4 x + 2 y - 42

解答ステップ

\frac{\partial}{\partial x} (x^{2} + y^{2} + (21 - x - y)^{2})

簡素化

x^{2} + y^{2} + (21 - x - y)^{2} : 2 x^{2} + 2 x y - 42 x + 2 y^{2} - 42 y + 441

= \frac{\partial}{\partial x} (2 x^{2} + 2 x y - 42 x + 2 y^{2} - 42 y + 441)

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} (2 x^{2}) + \frac{\partial}{\partial x} (2 x y) - \frac{\partial}{\partial x} (42 x) + \frac{\partial}{\partial x} (2 y^{2}) - \frac{\partial}{\partial x} (42 y) + \frac{\partial}{\partial x} (441)

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

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

\frac{\partial}{\partial x} (42 x) = 42

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

\frac{\partial}{\partial x} (42 y) = 0

\frac{\partial}{\partial x} (441) = 0

= 4 x + 2 y - 42 + 0 - 0 + 0

簡素化

= 4 x + 2 y - 42