(\partial)/(\partial y)(x^2+xy+y^2+y)

解

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

x + 2 y + 1

解答ステップ

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

x

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

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

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

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

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

= 0 + x + 2 y + 1

簡素化

= x + 2 y + 1