Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 27

Find an equation of the tangent to the curve $x^2+y^2 = (2x^2+2y62-x)^2$ at the point $\displaystyle \left(0,\frac{1}{2}\right)$ using Implicit Differentiation.
If $y'= m\text{ (slope)}$ then,
$\displaystyle \frac{d}{dx} + (x^2) \frac{d}{dx} (y^2) = \frac{d}{dx} (2x^2 + 2y^2 - x)^2$


$
\begin{equation}
\begin{aligned}
2x + 2y \frac{dy}{dx} & = 2 (2x^2+2y^2-x) \frac{d}{dx}(2x^2+2y^2-x)\\
\\
2x + 2y \frac{dy}{dx} & = 2 (2x^2+2y^2-x) \left[ 2 \frac{d}{dx} (x^2) + 2 \frac{d}{dx} (y^2) - \frac{d}{dx} (x) \right]\\
\\
2x + 2y \frac{dy}{dx} & = 2 (2x^2 + 2y^2 - x) \left[ (2)(2x) + (2)(2y) \frac{dy}{dx} - 1\right]\\
\\
2x + 2y \frac{dy}{dx} & = (4x^2 + 4y^2 - 2x) \left(4x +4y \frac{dy}{dx} -1 \right)\\
\\
2x + 2y \frac{dy}{dx} & = 16x^3+16x^2y \frac{dy}{dx} - 4x^2 +16xy^2 + 16y^3 \frac{dy}{dx} - 4y^2 - 8x^2 - 8xy \frac{dy}{dx} + 2x
\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}
2y \frac{dy}{dx} - 16x^2y \frac{dy}{dx} - 16y^3 \frac{dy}{dx} + 8xy \frac{dy}{dx} &= 16x^3 - 4x^2 + 16xy^2 - 4y^2 - 8x^2 + \cancel{2x} - \cancel{2x}\\
\\
\frac{dy}{dx} ( 2y - 16x^2y - 16y^3 + 8xy) &= 16x^3 - 4x^2 + 16xy^2 - 4y^2 - 8x^2\\
\\
\frac{\frac{dy}{dx} \cancel{( 2y - 16x^2y - 16y^3 + 8xy)} }{\cancel{ 2y - 16x^2y - 16y^3 + 8xy}} &= \frac{ 16x^3 - 4x^2 + 16xy^2 - 4y^2 - 8x^2}{2y - 16x^2y - 16y^3 + 8xy}
\end{aligned}
\end{equation}
$


$\displaystyle \frac{dy}{dx} = \frac{ 16x^3 - 4x^2 + 16xy^2 - 4y^2 - 8x^2}{2y - 16x^2y - 16y^3 + 8xy} \qquad \text{ or } \qquad y' = \frac{ 16x^3 - 4x^2 + 16xy^2 - 4y^2 - 8x^2}{2y - 16x^2y - 16y^3 + 8xy} $

For $x= 0 $ and $\displaystyle y = \frac{1}{2}$, we obtain

$
\begin{equation}
\begin{aligned}
y' = m &= \frac{16(0)^3 - 4(0)^2 + 16(0)\left( \frac{1}{2}\right)^2 - 4 \left( \frac{1}{2} \right)^2 - 8 (0)^2}{2 \left( \frac{1}{2} \right) - 16 (0)^2 \left( \frac{1}{2} \right) - 16 \left( \frac{1}{2} \right)^3 + 8(0) \left( \frac{1}{2} \right)}\\
\\
m &= \frac{0-0+0-1-0}{1-0-2+0}\\
\\
m &= \frac{-1}{-1}\\
\\
m &= 1
\end{aligned}
\end{equation}
$

Using point slope form

$
\begin{equation}
\begin{aligned}
y - y_1 & = m (x-x_1)\\
\\
y - \frac{1}{2} & = 1 (x - 0)\\
\\
y - \frac{1}{2} &= x\\
\\
y & = x + \frac{1}{2} && \text{Equation of tangent line at } \left(0, \frac{1}{2} \right)
\end{aligned}
\end{equation}
$

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