derivative of (sqrt(x+y+1)/(x-1))

解

\frac{d}{d x} (\frac{x + y + 1}{x - 1})

\frac{x \frac{d y}{d x} - x - 2 y - \frac{d y}{d x} - 3}{2 ( x - 1 ) ^{2} x + y + 1}

解答ステップ

\frac{d}{d x} (\frac{x + y + 1}{x - 1})

商の法則を適用:

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

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

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

\frac{d}{d x} (x - 1) = 1

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

簡素化

\frac{\frac{\frac{d y}{d x} + 1}{2 x + y + 1} ( x - 1 ) - 1 \cdot x + y + 1}{( x - 1 ) ^{2}} : \frac{x \frac{d y}{d x} - x - 2 y - \frac{d y}{d x} - 3}{2 ( x - 1 ) ^{2} x + y + 1}

= \frac{x \frac{d y}{d x} - x - 2 y - \frac{d y}{d x} - 3}{2 ( x - 1 ) ^{2} x + y + 1}