I want to find the general formula which gives the rate of change of the area with respect to the circumference. (Of a circle)
I know that I can use differentiation formulas but I don't want to do it that way.
We have : $C=2\pi r$ when we isolate for r it gives : $r=\frac{C}{2\pi}$
and $A=\pi*r^2$
Basically what I did is this :
A=$\frac{(\pi*(r1+r2)^2)-(\pi*r1^2)}{r1+r2-r1}$
I'm supposed to obtain Circumference divided by $2*\pi$
What did I do wrong ?
Other than the missing limit, the problem is that you are setting up the computation for $\frac{\partial A}{\partial r}(r)$ and you are expecting that substituting $r = \frac{C}{2\pi}$ it will give you $\frac{\partial A}{\partial C}(C)$. This is not possible. Instead, by the chain rule we know that $$ \frac{\partial A}{\partial C}(C) = \frac{\partial r}{\partial C}(C) \; \frac{\partial A}{\partial r}(r(C)) = \frac{1}{2\pi} \; \frac{\partial A}{\partial r}\left(\frac{C}{2\pi}\right). $$
To compute $\frac{\partial A}{\partial C}(C)$ directly as a limit, consider $h > 0$. Then $$ \begin{align} \frac{\partial A}{\partial C}(C) &= \lim_{h \to 0} \frac{\pi \left(\frac{C+h}{2\pi}\right)^2 - \pi \left(\frac{C}{2\pi}\right)^2}{h} \\ &= \lim_{h \to 0} \frac{(C+h)^2 - C^2}{4\pi h}\\ &= \lim_{h \to 0} \frac{2Ch + h^2}{4\pi h}\\ &= \underbrace{\left(\lim_{h \to 0} \frac{2Ch}{4\pi h}\right)}_{C/2\pi} + \underbrace{\left(\lim_{h \to 0} \frac{h^2}{4\pi h}\right)}_0. \end{align} $$
On the other hand, if you know that $\frac{\partial}{\partial x} x^2 = 2x$ it is way quicker to just observe that $$ A = \pi r^2 = \frac{1}{4\pi} C^2 $$ immediately implies $$ \frac{\partial A}{\partial C}(C) = \frac{1}{4 \pi} \; \frac{\partial}{\partial C} C^2 = \frac{1}{2\pi} C. $$