Implicit derivative of $x^2+y^2=(2x^2+2y^2-x)^2$

Question

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:)

Answer 1

Your answer seems true: $$2x+2yy'=2(2x^2+2y^2-x)(4x+4yy'-1),$$ but there is a mistake in your nominator.

It should be $$y'=\frac{2(-4x^3+3x^2+y^2-4xy^2)}{y(8y^2+8x^2-4x-1)}.$$

Answer 2

$$x^2+y^2=(2x^2+2y^2-x)^2$$ 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; $$y'=\dfrac 2y\dfrac {(4x^3+4y^2x-3x^2-y^2)}{(1-4(2x^2+2y^2-x))}$$

Answer 3

An alternative way. Set $Y=x^2+y^2$, so $$ F(x,y)=x^2+y^2-(2x^2+2y^2-x)^2=Y-(2Y-x)^2=G(x,Y) $$ then $$ Y'=-\frac{G_x}{G_Y}=-\frac{2(2Y-x)}{1-4(2Y-x)} $$ and $$ Y'=2x+2yy'\quad\implies\quad y'=\frac{Y'-2x}{2y} $$ then substitute $Y'$ in the last expression and, furthermore, substitute $Y=x^2+y^2$.

Answer 4

This is intended as a comment, rather than an answer, but is too long to attach to the posted problem. The given curve equation is the Cartesian form for a cardioid, which is why the expression is peculiar. (The equation and the derivative expression are far simpler in polar coordinates.) The implicit form for the derivative also has a strange feature.

The curve for $ \ x^2+y^2 \ = \ (2x^2+2y^2-x)^2 \ $ is marked in blue in the graph below. If we examine the factors in the derivative expression,

$$ y' \ \ = \ \ \frac{2·(-4x^3 \ + \ 3x^2 \ + \ y^2 \ - \ 4xy^2)}{y·(8y^2 \ + \ 8x^2 \ - \ 4x \ - \ 1)} \ \ , $$

the green curve is the locus of points at which the factor in parentheses in the numerator equals zero, the orange curve, that for which the factor in parentheses in the denominator is zero. So the cardioid has horizontal tangents at the intersections of the blue and green curves and vertical tangents where the orange and blue curves meet.

The denominator is also equal to zero for $$ y \ = \ 0 \ \ \Rightarrow \ \ x^2 \ = \ (2x^2 - x)^2 \ \ \Rightarrow \ \ 4x^4 \ - \ 4x^3 \ \ = \ \ 0 \ \ \Rightarrow \ \ 0 \ , \ 1 \ \ . $$ So there is another vertical tangent at $ \ ( 1 \ , \ 0 ) \ \ . $ However, if we insert $ \ x \ = \ 0 \ \ , \ \ y \ = \ 0 \ \ $ into the expression, we find that the derivative is indeterminate, which is a sign that there is something odd about the curve there. We see in the graph that the origin is the location of a "cusp" in the curve, where the first derivative changes continuously through zero, but the second derivative is discontinuous.