(\partial)/(\partial y)(ln(x^2+1)+y^2)

解

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

2 y

解答ステップ

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

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

\frac{\partial}{\partial y} (ln (x^{2} + 1)) = 0

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

= 0 + 2 y

簡素化

= 2 y