(\partial)/(\partial x)((xy)/(x+y))

解答

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

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

求解步骤

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

将

y

视为常数

将常数提出:

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

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

使用微分除法定则:

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

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

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

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

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

化简

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

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

d er i v a t i v e f (x) = ln (6 - x^{2})

\frac{\partial}{\partial x} (x cos (y) - y sin (x))

x \to 1 lim (\frac{( 5 - 5 x )}{1 - x})

x \frac{d y}{d x} - y = 0, y (a) = b

y^{^{'}} + 2 x y = x