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

解

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

\frac{1}{y + 1}

解答ステップ

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

y

を定数として扱う

定数を除去:

(a \cdot f)^{'} = a \cdot f^{'}

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

和/差の法則を適用:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

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

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

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

簡素化

= \frac{1}{y + 1}