Differentiation of the Unit Circle

Here are two solutions:

Solution 1: If you start with $x^{2}+y^{2}=1$ , taking the derivative of both sides with respect to $x$ results in:

$\frac{d}{dx}(x^{2}+y^{2})=\frac{d}{dx}(1)$

$2x+2y\frac{dy}{dx}=0$

$2y\frac{dy}{dx}=-2x$

$\frac{dy}{dx}=-\frac{x}{y}$ .

$x^{2}+y^{2}=1$ 's graph in a $2$ -dimensional Cartesian coordinate system with coordinate axes $x$ and $y$ is the boundary of a circle centered at the origin with radius $1$ .

Let $r$ represent that radius.

Therefore, substituting $x=r\cos \theta$ , $y=r\sin \theta$ , and $r=1$ into $\frac{dy}{dx}=-\frac{x}{y}$ we obtain $\frac{dy}{dx}=-\frac{\cos \theta}{\sin \theta}$ , which simplifies to $\boxed{\frac{dy}{dx}=-\cot \theta}$ .

Solution 2:

This assumes a $2$ -dimensional Cartesian coordinate system with coordinate axes $x$ and $y$ .

The slope of line $\omega$ is $\tan \theta$ , by the definition of the unit circle.

Line $\psi$ is perpendicular to line $\omega$ , and, therefore, by the theorem that states that perpendicular lines have negative reciprocal slopes,

The slope of line $\psi$ is $-\frac{1}{\tan \theta}$ , which is equal to $-\cot \theta$ .

By definition, $\frac{dy}{dx}$ is the slope of the tangent line of the graph of the relation between $x$ and $y$ at any $x$ value.

Since line $\psi$ is the tangent line of the relation $x^{2}+y^{2}=1$ , its slope is $\frac{dy}{dx}$ .

Therefore, $\boxed{\frac{dy}{dx}=-\cot \theta}$ .

$x=r\cos \theta, y=r\sin \theta, \dfrac{dy}{dx}=-\dfrac{x}{y}=-\dfrac{r\cos \theta}{r\sin \theta}=\boxed {-\cot \theta}$