What is (\partial)/(\partial x)((2xy)/(x-y)) ?

The answer to (\partial)/(\partial x)((2xy)/(x-y)) is -(2y^2)/((x-y)^2)

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(\partial)/(\partial x)((2xy)/(x-y))

\frac{\partial}{\partial x} (\frac{2 x y}{x - y})

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

Solution steps

\frac{\partial}{\partial x} (\frac{2 x y}{x - y})

Treat

y

as a constant

Take the constant out:

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

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

Apply the Quotient Rule:

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

= 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

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

Simplify

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

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

f (x) = \frac{x}{7}

y = - x^{2} - 8 x - 12, y = 0, [- 4, - 1]

f (x) = e^{- 8 x}

\int_{0}^{3} x^{2} - 6 d x

\int_{0}^{π} 13 sin^{2} (4 x) d x

What is (\partial)/(\partial x)((2xy)/(x-y)) ?
The answer to (\partial)/(\partial x)((2xy)/(x-y)) is -(2y^2)/((x-y)^2)