What is the implicit derivative of #y=(x-y)^2+4xy+5y^2 #?

1 Answer
Krishna V.
May 29, 2017

The implicit derivative is #dy/dx= (2x+2y)/(-12y-2x+1)#.

Explanation:

First, you would derive this by turning #y# into #dy/dx# when needed since it is a function while treating #x# as the variable. For the #(x-y)^2#, the chain rule is applied. This will result in:

#dy/dx=2(x-y)*(1-dy/dx)+4y+4xdy/dx+10ydy/dx#

For simplicity, we can distribute the values in the parentheses and then implement quadratic multiplication.

#dy/dx=(2x-2y)*(1-dy/dx)+4y+4xdy/dx+10ydy/dx#

#dy/dx=2x-2xdy/dx-2y+2ydy/dx+4y+4xdy/dx+10ydy/dx#

We now simplify like terms to get:

#dy/dx=2x+12ydy/dx+2y+2xdy/dx#

Move over all of the terms with #dy/dx# to the other side:

#-12ydy/dx-2xdy/dx+dy/dx=2x+2y#

Factor the #dy/dx# from the left side to get:

#dy/dx(-12y-2x+1)=2x+2y#

Divide the left and right sides by #(-12y-2x+1)# to achieve:

#dy/dx= (2x+2y)/(-12y-2x+1)#

There is your answer! Hope it helps!

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