How do you differentiate #y=(x-y)^2/(x+y)#?

Answer 1

Given: #y=(x-y)^2/(x+y); x!=-y#

Multiply both sides by #x+y#:

#xy+y^2 = (x-y)^2#

Expand the square:

#xy+y^2 = x^2 -2xy+y^2#

Combine like terms:

#3xy = x^2#

Divide both sides by #3x#:

#y = 1/3x#

#dy/dx = 1/3#

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

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Answer 2

To differentiate the function ( y = \frac{{(x - y)^2}}{{x + y}} ) with respect to ( x ), we'll use the quotient rule and implicit differentiation.

Let ( u = (x - y)^2 ) and ( v = x + y ). Then,

( \frac{{dy}}{{dx}} = \frac{{v \frac{{du}}{{dx}} - u \frac{{dv}}{{dx}}}}{{v^2}} )

First, differentiate ( u ) and ( v ) with respect to ( x ):

( \frac{{du}}{{dx}} = 2(x - y)(1 - \frac{{dy}}{{dx}}) - 1 )

( \frac{{dv}}{{dx}} = 1 + \frac{{dy}}{{dx}} )

Now, substitute these into the quotient rule formula:

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

Now, solve for ( \frac{{dy}}{{dx}} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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