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

解

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

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

解答ステップ

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

x

を定数として扱う

商の法則を適用:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

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

= \frac{1 \cdot ( 1 + y ^{2} ) - 2 y ( x + y )}{( 1 + y ^{2} ) ^{2}}

簡素化

\frac{1 \cdot ( 1 + y ^{2} ) - 2 y ( x + y )}{( 1 + y ^{2} ) ^{2}} : \frac{- y ^{2} - 2 x y + 1}{( 1 + y ^{2} ) ^{2}}

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