Relationship between tan(x) and derivative of Unit Circle

Question

Can someone show the relationship between the function:

$\tan x$

and the derivative of a circle

$\pm {x\over \sqrt{1-x^2}}$

I read this article but I was not able to connect the two functions through the method described. Is this something just not worth attempting? Relationship Between Tangent Function and Derivative

Answer 1

As also Zen answered regarding the post you mentioned, taking $\tan$ of an angle is equivalent to the slope or technically derivative of the line.

First, we need to find the slope of the radius line:

$\tan \theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{\sqrt{1-\cos^2 \theta}}{{\cos \theta}}=\dfrac{\sqrt{1-x^2}}{x}$

The tangent line is perpendicular to the radius. Thus we should inverse slope of radius line, and multiply by $-1$ to find the slope of the tangent line. Note that $x=\cos\theta$.

Now the tangent slope is:

$\dfrac{dy}{dx}=-\dfrac{1}{\tan(\theta)}=-\dfrac{x}{\sqrt{1-x^2}}$

In other words:

$\dfrac{dy}{dx}=\dfrac{d(\sin\theta)}{d(\cos\theta)}=\dfrac{\cos\theta d\theta}{-\sin\theta d\theta}=-\dfrac{1}{\tan \theta}=-\dfrac{x}{\sqrt{1-x^2}}$