For the area of the circle, it is given as a function of the radius. The area of the square is given as a function of its side lengths. Radii and side lengths are inherently different things, so taking the derivative with respect to them should certainly show different behavior.
You can reformulate the area of a square, too, to have the same behavior. Writing it as a function of something other than its side length. Observe:
Take a square of side length x, so its area is \(\displaystyle A(x)=x^2\). Let's create a new variable \(\displaystyle s=x/2\) (half the side of the square... let's pretend it is a kind of "radius" for the square. It is, in fact, the radius of an inscribed circle). Now the area of the original square is given by \(\displaystyle A(s)=4s^2\), and its Perimiter is \(\displaystyle P(s)=8s\). Note that \(\displaystyle dA/ds = P(s)\).
For the circle, let \(\displaystyle x=\sqrt{\pi} r\), and let's call this a "side" of the circle. Then the area of the circle is \(\displaystyle A(x) = x^2\), and \(\displaystyle C(x) =2\sqrt{\pi}x\). Then \(\displaystyle dA/dx \neq C(x)\).
Hopefully this shows you that it depends on how the variables are constructed.