What is the d/(dy)((x-y)/(x+y)) ?

The d/(dy)((x-y)/(x+y)) is -(2x)/((x+y)^2)

What is the first d/(dy)((x-y)/(x+y)) ?

The first d/(dy)((x-y)/(x+y)) is -(2x)/((x+y)^2)

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

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

Solution steps

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

Treat

x

as a constant

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

x \to 0 lim (\frac{e ^{x} - 1}{x ^{9}})

x^{2} y^{^{''}} - 2 x y^{^{'}} + 2 y = 0

f (x) = - x^{2} + 6 x - 8, 4, - 2

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

\int_{0}^{6} t e^{- t} d t

What is the d/(dy)((x-y)/(x+y)) ?
The d/(dy)((x-y)/(x+y)) is -(2x)/((x+y)^2)
What is the first d/(dy)((x-y)/(x+y)) ?
The first d/(dy)((x-y)/(x+y)) is -(2x)/((x+y)^2)