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Use implicit differentiation to find an equation of the tangent line to the curve: $x^2+y^2=(2x^2+2y^2-x)^2$

At the point: $(0, \frac{1}{2})$

Hi I'm really lost with this question, can somebody please work through it for me so I have a example for my other questions. Thanks

Chris Brooks
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user142105
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1 Answers1

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Differentiating both sides, we get:
$2x+2yy'=2(2x^2+2y^2-x)*(4x+4yy'-1)$.
Plugging in $x=0, y=\frac 12$, we get:
$y'=2y'-1$,
or $y'=1$.
Therefore, $y-\frac 12=y'(0)(x-0)=x$. This will be transformed into:
$y=x+\frac 12$, which is the equation for tangent line at $x=0$.

pxc3110
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