How do you differentiate $y = \frac{{(x - y)}^{2}}{x + y}$ $y=(x-y)^2/(x+y)$ ?

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How do you differentiate $y = \frac{{(x - y)}^{2}}{x + y}$ $y=(x-y)^2/(x+y)$ ?

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Answer 1 · 2017-09-25T18:19:33

Given: $y = \frac{{(x - y)}^{2}}{x + y}; x \neq - y$ $y=(x-y)^2/(x+y); x!=-y$

Multiply both sides by $x + y$ $x+y$ :

$x y + y^{2} = {(x - y)}^{2}$ $xy+y^2 = (x-y)^2$

Expand the square:

$x y + y^{2} = x^{2} - 2 x y + y^{2}$ $xy+y^2 = x^2 -2xy+y^2$

Combine like terms:

$3 x y = x^{2}$ $3xy = x^2$

Divide both sides by $3 x$ $3x$ :

$y = \frac{1}{3} x$ $y = 1/3x$

$\frac{d y}{d x} = \frac{1}{3}$ $dy/dx = 1/3$

If you do not believe the result, here is a graph of $y = \frac{{(x - y)}^{2}}{x + y}$ $y=(x-y)^2/(x+y)$ to prove that is merely a line with a slope of $\frac{1}{3}$ $1/3$ :

![www.desmos.com/calculator]( useruploads.socratic.org )

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How do you differentiate $y = \frac{{(x - y)}^{2}}{x + y}$ $y=(x-y)^2/(x+y)$ ?

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