<s>[INST] Implicit derivative of $x^2+y^2=(2x^2+2y^2-x)^2$ Hi there.
I have the following function and want to calculate $y'$:
$x^2+y^2=(2x^2+2y^2-x)^2$
I've used implicit differentiation to solve it and my answer is:
$y'=\frac{(-4x^3+3x^2-4xy^2+y^2)}{y(8x^2-4x+8y^2-1)}$
However, my calculation gets huge ( I use the chainrule) and I'm just curious if there's a smarter/more simple way to calculate this?
Thank you:)
 [/INST] $$x^2+y^2=(2x^2+2y^2-x)^2$$
Differentiate with respect to the variable $x$:
$$2x+2yy'=2(2x^2+2y^2-x)(4x+4yy'-1)$$
$$x+yy'=(2x^2+2y^2-x)(4x+4yy'-1)$$
$$(x+yy')(1-4(2x^2+2y^2-x))=-(2x^2+2y^2-x)$$
$$(x+yy')=-\dfrac {(2x^2+2y^2-x)}{(1-4(2x^2+2y^2-x))}$$
Finally we get;
$$y'=\dfrac 2y\dfrac {(4x^3+4y^2x-3x^2-y^2)}{(1-4(2x^2+2y^2-x))}$$
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