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!

Related questions
See all questions in Implicit Differentiation
Impact of this question
1916 views around the world
You can reuse this answer
Creative Commons License