How can I graph the derivative of 1/4th of a circle or a semicircle in a piecewise function? (Also other kinds of piecewise functions) 

I'm having trouble with questions like these. In the first image, the original function is what is the two sharp lines and a semicircle in between. I understand how to find and graph the derivative of the parts that are straight lines however i don't understand how to find the derivative of the semicircle. I can see that the radius is 2 so
$x^2+y^2 = 2^2 = 4$
Do I implicitly differentiate with respect to $x$ in order to get $y'$ and the graph that?
If i do that I get $2x+2yy' = 0 \implies y' = \frac{-x}{y}$
Now what? Also this is a semicircle so must I differentiate $\frac{x^2+y^2}{2} = 2$?
Similarly, in the second question I have 1/4th of a circle, how can I take the derivative of that and graph it. Finally, in the second question, I understand that the derivative of a parabola would be a linear function because (for example) the derivative of $x^2$ is $2x$ but how can I graph this with just the information that it is a parabola and I'm not given the function itself. I know that if the function is decreasing then the derivative must be negative. I understand intuitively why it must be linear but is there any graphing "rule" which would make me know this?
EDIT: is it as simple as (for the first question) that the function is decreasing so the derivative will be below the x-axis and for the second question that the inflection point is at $x=2$ so it is decreasing from 0 to 2 so function will be negative and increasing from 2 to 4 so it will be positive there? But how does it get that shape that makes it look like an $x^3$ graph? it could look like a line, or a parabola or anything, why does it have that specific shape?
 A: I don't think you are expected to provide exact graphs, but there is more information available from the graphs than you are using.
In your problem with the half-circle and the two straight lines, note first of all that it is a half-circle, not a full circle. Also note that the center is $(2,0)$, not $(0,0)$. The equation of the full circle is $(x - 2)^2 + y^2 = 4$, and we can solve this for $y$:
$$y = -\sqrt{4 - (x - 2)^2} = -\sqrt{4x - x^2}$$
where I've chosen the negative square root because your graph is of the bottom half of the circle, where $y < 0$. The derivative is given by $$y' = \dfrac{x - 2}{\sqrt{4x-x^2}}$$
Some things to observe about this: At $x = 2, y' = 0$. Near $x = 0$, the numerator is close to $-2$ while the denominator goes to $0$ from above, so $y' \to -\infty$. Similarly at $x = 4$, the numerator is near $4$ as the denominator goes to $0$ from above, so $y' \to +\infty$.
You don't actually need to find the derivative formula to spot those three facts: At $x = 2$ is the minimum of the curve, and at minimums the derivative is always $0$. It is also obvious from the graph that the tangent line there must be horizontal. At $x = 0$ and $x = 4$, the tangents to the circle are vertical, so $y' = \pm\infty$. Near $0$, the tangents decrease as you travel right, so their slope is negative. Hence $y' \to -\infty$ at $x = 0$ and similarly you can see that $y' \to +\infty$ at $x = 4$.
Another thing we can see is that the slope is increasing. Looking at the circle, as we move right, the tangents become less and less negative in their slope until we get to $x = 2$ where it is $0$. After which they become more and more positive until they are vertical again at $x = 4$. So the graph of $y'$ has to be increasing.
These facts alone necessitate a graph shaped liked the one shown. Except note that it has asymptotes at $x = 0$ and $x = 4$. It is strictly vertical there. $y = (x - 2)$ and even $y = (x-2)^3$ do not work, as neither has asymptotes. Because the graph of the derivative crosses the $x$-axis smoothly at $x = 2$, it will do so at some slope. But to the left, it has curve down towards vertical as it approaches $x = 0$, and to to the right, it has to curve up towards vertical as it approaches $x = 4$.
The only thing more that would really be useful here is to know what slope it crosses the $x$-axis at. For that you need to calculate the 2nd derivative and evaluate it at $x = 2$. I have no shortcuts for that, but the result is $\frac 12$. This means the graph in your picture is somewhat inaccurate, as its slope appears to be about $1$, but this is a minor correction.

For the other graph, the 1/4 circle is very much the same, but you are only looking at one side. Look at its graph (which doesn't look quite circular because the horizontal and vertical scales are not the same). Again, at $x = -6$, the tangent to the circle is going to be horizontal. As you move right, the tangents start to slope down more and more until at $x = -2$, the tangent becomes vertical. Hence the derivative will be $0$ at $x = -6$, become more and more negative as you move right until it has an asymptote ($y' \to -\infty$) at $x = -2$. Because it has some finite slope (actually, $-1/2$ again) when crossing the $x$-axis at $x = -6$, but has an asymptote at $x = -2$, it has to curve down as the answer shows.
On the parabola side, yes it has to be linear, but you can figure out what line from what you know about about parabolas. You know that the vertex is at (4,6) and it opens down, so it has an equation of the form $$y = 6 - a(x -4)^2$$We also see that it passes through $(9,0)$, so $0 = 6 - a(9-4)^2$which gives $a = \frac6{25}$, and therefore $$y' = -\frac {12}{25}(x - 4)$$
That is, the derivative is going to have a slope that is close to $-\frac 12$, and will be $0$ at $x = 4$ (which you already knew, since the vertex is there). So its graph is going to be $0$ at $x=4$ and extend to the right at a slope near to $-\frac 12$.
Note that the answer once again was not completely accurate in the slope. You were not expected to produce just the right slope. Simply graphing a moderately negative slope from $(4,0)$ to the right would have been enough.
