Graphing dy/dx of a parametric equation.

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Consider the first quadrant of a circle. We can represent the first quadrant of a circle as:

$y_1 = \sqrt{1-x^2},$ such that $0\leq x \leq 1 \\$

and in parametric terms as:

$\left(\frac{1-t^{2}}{1+t^{2}},\frac{2t}{1+t^{2}}\right)$.

Let us compute the derivatives for both. So:

$\frac{dy_1}{dx} = \sqrt{1-x^2},$

and for the parametric equation:

$\frac{dy_2}{dx} = \left( \frac{2(1+t^2) - 2t(2t)}{-2t(1+t^2) - 2t(1-t^2)} \right)$.

What I'm confused about is how we would represent this accurately. Clearly, we cannot simply make t=x and then graph. We could replace t by reordering the x parameter like so:

$t = \sqrt{\frac{1-x}{1+x}}$

But, even then, $\frac{dy_1}{dx} ≠ \frac{dy_2}{dx}$. So how exactly are we supposed to graph $\frac{dy_2}{dx}$?