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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\frac{d}{d x} (sin (cos (4 x)))

\frac{1}{5 x ^{\frac{4}{5}}}

f (x) = \frac{x ^{3} + 1}{x ^{3} - 1}

x^{3} y^{^{'''}} + x y^{^{'}} - y = 0

\int e^{- 1}

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