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

解

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

2 y

解答ステップ

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

x, z

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

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

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

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

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

= 0 + 2 y + 0 - 0

簡素化

= 2 y