What is the implicit derivative of #4= (x+y)^2 -3xy#?

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What is the implicit derivative of #4= (x+y)^2 -3xy#?

↳Redirected from "Suppose that I don't have a formula for #g(x)# but I know that #g(1) = 3# and #g'(x) = sqrt(x^2+15)# for all x. How do I use a linear approximation to estimate #g(0.9)# and #g(1.1)#?"

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Answer 1 · 2016-01-05T15:20:11

I found: #(dy)/(dx)=(-2x+y)/(2y-x)#

Explanation:

You need to derive #y# as well as it represents a function of #x#.
So, for example, if you have #y^2# the derivative will be #2y(dy)/(dx)#.
In your case:
#0=2(x+y)*(1+1(dy)/(dx))-3y-3x(dy)/(dx)#
#0=2x+2x(dy)/(dx)+2y+2y(dy)/(dx)-3y-3x(dy)/(dx)#
We collect #(dy)/(dx)#:
#(dy)/(dx)=(-2x+y)/(2y-x)#

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What is the implicit derivative of #4= (x+y)^2 -3xy#?

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