derivative of (e^x/(e^x+e^y))

解

\frac{d}{d x} (\frac{e ^{x}}{e ^{x} + e ^{y}})

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

解答ステップ

\frac{d}{d x} (\frac{e ^{x}}{e ^{x} + e ^{y}})

商の法則を適用:

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

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

\frac{d}{d x} (e^{x}) = e^{x}

\frac{d}{d x} (e^{x} + e^{y}) = e^{x} + e^{y} \frac{d y}{d x}

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

拡張

e^{x} (e^{x} + e^{y}) - (e^{x} + e^{y} \frac{d y}{d x}) e^{x} : e^{x + y} - e^{x + y} \frac{d y}{d x}

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