I had the following equation for $f(r)$. I have to find the derivative of it. I think it will be pretty lengthy if I find derivate of individual terms, so I took summation ($\sum$) but I want to remove the summation sign at the end. Following is $f(r)$ followed by my solution for $\frac{d f(r)}{dr}$
$$\small f(r) = 1-\Big[ e^{-\lambda \pi r^2} + \lambda \pi r^2e^{-\lambda \pi r^2} + \frac{(\lambda \pi r^2)^2e^{-\lambda \pi r^2}}{2!} + \frac{(\lambda \pi r^2)^3e^{-\lambda \pi r^2}}{3!}... + ... \frac{(\lambda \pi r^2)^{(k-1)}e^{-\lambda \pi r^2}}{(k-1)!}\Big]$$
My Solution
First of all, $$f(r) = 1-\Big[\sum_{i=1}^{k-1} \frac{(\pi \lambda r^2)^{i-1} e^{-\pi \lambda r^2}}{(i-1)!}\Big]$$
$$\frac{d\ f(r)}{dr} = -\frac{d}{dr}\Big[\sum_{i=1}^{k-1} \frac{(\pi \lambda r^2)^{i-1} e^{-\pi \lambda r^2}}{(i-1)!}\Big]$$
$$\frac{d\ f(r)}{dr} = -\sum_{i=1}^{k-1} \frac{(\pi \lambda)^{i-1}}{(i-1)!}\frac{d}{dr}r^{2(i-1)} e^{-\pi \lambda r^2}$$
$$ = -\sum_{i=1}^{k-1} \frac{(\pi \lambda)^{i-1}}{(i-1)!} \Big[2(i-1)r^{2(i-1)-1}e^{\pi\lambda r^2} + r^{2(i-1)}(2\pi\lambda r)e^{\pi \lambda r^2}\Big] $$
$$ = -\sum_{i=1}^{k-1} \frac{(\pi \lambda)^{i-1}e^{\pi \lambda r^2} r^{2(i-1)}}{(i-1)!} \Big[2(i-1)r^{-1} + 2\pi\lambda r\Big] $$
It has become very complicated. I want the final answer which is actually as follows: (I don't know how to remove the summation sign)
Actual Answer: (which I cannot get)
$\frac{d\ f(r)}{dr} = \frac{2(\lambda\pi)^k}{(k-1)!}r^{2k-1}e^{-\lambda \pi r^2}$, for $k\geq 1$
You have $$\newcommand{\la}{\lambda}f(r)=1-g(e^{\la\pi r^2})$$ where $$g(x)=e^{-x}\sum_{i=0}^{k-1}\frac{x^i}{i!}.$$ Then $$g'(x)=-e^{-x}\sum_{i=0}^{k-1}\frac{x^i}{i!} +e^{-x}\sum_{i=1}^{k-1}\frac{x^{i-1}}{(i-1)!}=-e^{-x}\frac{x^{k-1}}{(k-1)!}.$$ Therefore $$f'(r)=-2\la\pi r g'(e^{\la\pi r^2}) =2\la\pi r(\la\pi r^2)^{k-1}\frac{e^{-\la\pi r^2}}{(k-1)!}$$ etc.