Question

use implicit differentiation to find an equation of the tangent line to the curve at the given point.
$x^{2}+y^{2}=(5x^{2}+4y^{2}-x)^{2}$, $(0,\frac{1}{4})$ (cardioid)

Explanation:

Step1: Differentiate both sides

Differentiate $x^{2}+y^{2}=(5x^{2}+4y^{2}-x)^{2}$ with respect to $x$.
Using the chain - rule and sum - rule of differentiation.
For the left - hand side: $\frac{d}{dx}(x^{2}+y^{2}) = 2x + 2y\frac{dy}{dx}$.
For the right - hand side: Let $u = 5x^{2}+4y^{2}-x$, then $(5x^{2}+4y^{2}-x)^{2}=u^{2}$.
$\frac{d}{dx}(u^{2})=2u\frac{du}{dx}$, and $\frac{du}{dx}=(10x + 8y\frac{dy}{dx}-1)$.
So $\frac{d}{dx}(5x^{2}+4y^{2}-x)^{2}=2(5x^{2}+4y^{2}-x)(10x + 8y\frac{dy}{dx}-1)$.
We get $2x + 2y\frac{dy}{dx}=2(5x^{2}+4y^{2}-x)(10x + 8y\frac{dy}{dx}-1)$.

Step2: Substitute the point $(0,\frac{1}{4})$

Substitute $x = 0$ and $y=\frac{1}{4}$ into $2x + 2y\frac{dy}{dx}=2(5x^{2}+4y^{2}-x)(10x + 8y\frac{dy}{dx}-1)$.
Left - hand side: $2(0)+2(\frac{1}{4})\frac{dy}{dx}=\frac{1}{2}\frac{dy}{dx}$.
Right - hand side: First, when $x = 0$ and $y=\frac{1}{4}$, $5x^{2}+4y^{2}-x=4(\frac{1}{4})^{2}=\frac{1}{4}$.
Then $2(5x^{2}+4y^{2}-x)(10x + 8y\frac{dy}{dx}-1)=2(\frac{1}{4})(8(\frac{1}{4})\frac{dy}{dx}-1)=\frac{1}{2}(2\frac{dy}{dx}-1)$.
So $\frac{1}{2}\frac{dy}{dx}=\frac{1}{2}(2\frac{dy}{dx}-1)$.
$\frac{1}{2}\frac{dy}{dx}=\frac{1}{2}\times2\frac{dy}{dx}-\frac{1}{2}$.
$\frac{1}{2}\frac{dy}{dx}= \frac{dy}{dx}-\frac{1}{2}$.
$\frac{dy}{dx}-\frac{1}{2}\frac{dy}{dx}=\frac{1}{2}$, $\frac{1}{2}\frac{dy}{dx}=\frac{1}{2}$, $\frac{dy}{dx}=1$.

Step3: Find the equation of the tangent line

The equation of a tangent line in point - slope form is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(0,\frac{1}{4})$ and $m = 1$.
$y-\frac{1}{4}=1(x - 0)$.
$y=x+\frac{1}{4}$.

Answer:

$y=x+\frac{1}{4}$