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

解答

\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}

求解步骤

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

使用微分除法定则:

(\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}}

化简

\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}

\int_{0}^{2 - y} \frac{5}{8} x^{3} d x

\frac{d}{d x} (cos (2 x) sin (\frac{x}{3}))

x \to + 4 lim (\frac{3 x}{4} + 2)

\frac{\partial}{\partial x} (2 e^{x^{8} y})

x \to 0 lim (x cos (\frac{9}{x}))