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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

f (x) = x^{2} - 2

2 \frac{d y}{d x} = x y (x + 1)

\int x (2 x + 7)^{8} d x

\int \frac{1}{x ( 2 - x )} d x

\int \frac{1}{6 x - 1} d x

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