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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

= 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}}

= 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}}

Simplify

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

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

\int_{\frac{1}{2}}^{5} 14 x ln (2 x) d x

(3 x^{2} y + x y^{2}) d x + (x^{3} + x^{2} y) d y = 0

\int_{5}^{6} \frac{x}{x ^{2} + 4 x + 20} d x

y^{^{''}} + 4 y^{^{'}} + 5 y = e^{x}

\frac{d}{d x} (- 8 e^{- 8 x})

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