Why does the derivative equation of the unit semicircle equation intersect the semicircle at $x=-\frac{1}{\phi}$? 
I was playing around in desmos and I discovered something interesting, but I am nowhere near advanced enough to tackle this.  
The circle equation $f(x)=\sqrt{1-x^2}$
and its derivative (I don't know how to find the equation - if possible could you explain this as well?)   
intersect at or extremely close to (as in I zoomed in as far as I could in desmos) the point where x is equal to $\frac{-1}{\phi}$ where $\phi$ is 1.6180339...
I don't know whether all 3 lines actually cross at this point, but it seems like it's too perfect to be false. Is anybody able to come up with a reason why this happens?
 A: Expanding on Peter's comments, the derivative of $f(x)=\sqrt{1-x^2}$ can indeed be found using the chain rule. For more background on derivatives if you'd like it, see here. The chain rule states that, for two functions $g(x)$ and $h(x)$, the derivative of their composition $g(h(x))$ is given by
$$\big[g(h(x))\big]'=g'(h(x))\cdot h'(x)$$
In this case, we have $g(x)=\sqrt{x}$ and $h(x)=1-x^2$. Applying the chain rule, we have
\begin{align*}
g'(x)&=(x^{1/2})' \\
&=\frac{1}{2}x^{-1/2} \\
&=\frac{1}{2\sqrt{x}} \\
\implies g'(h(x))&=\frac{1}{2\sqrt{1-x^2}} \\
h'(x)&=(1-x^2)' \\
&=-2x
\end{align*}
Pulling these derivatives together, we find that
\begin{align*}
f'(x)&=\big[g(h(x))\big]' \\
&=g'(h(x))\cdot h'(x) \\
&=\frac{1}{2\sqrt{1-x^2}}(-2x) \\
&=\frac{-x}{\sqrt{1-x^2}}
\end{align*}

We can now set the function $f(x)$ equal to its derivative $f'(x)$ to see where the two intersect:
\begin{align*}
f(x)&=f'(x) \\[0.5ex]
\sqrt{1-x^2}&=\frac{-x}{\sqrt{1-x^2}} \\[0.5ex]
1-x^2&=-x
\end{align*}
Rearranging this, we end up with
$$x^2-x-1=0$$
This can be solved with the quadratic formula:
\begin{align*}
x&=\frac{1\pm\sqrt{1-4(-1)(1)}}{2} \\
&=\frac{1\pm\sqrt{5}}{2}
\end{align*}
Notice that this gives two solutions, whereas the two graphs you showed only intersect once. That's because the solution $x=\frac{1+\sqrt{5}}{2}$, when substituted back into $f(x)$, returns the square root of a negative number, which isn't a real solution. The solution $x=\frac{1-\sqrt{5}}{2}$ is therefore the only real solution. Even though the golden ratio, $\phi$, is normally given as $\phi=\frac{1+\sqrt{5}}{2}$, note that this solution we obtained is exactly $\frac{-1}{\phi}$:
$$\frac{-1}{\phi}=\frac{-2}{1+\sqrt{5}}\cdot\frac{1-\sqrt{5}}{1-\sqrt{5}}=\frac{-2(1-\sqrt{5})}{1-5}=\frac{-2(1-\sqrt{5})}{-4}=\frac{1-\sqrt{5}}{2}$$
This means that all three curves you graphed do cross at the same point! There might exist some kind of intuitive explanation as to why the intersection of a circle with its derivative involves the golden ratio, but I don't know what that explanation would be, unfortunately.
