I've been interested in working out what the derivative of $\Sigma$ is for a while now. I read somewhere that the derivative of $\Sigma$ is the sum of its derivatives. I wanted to reason my way through this, but what I got is different from what I'm told it is. Can anyone please help me understand where I went wrong and how to correct myself? (I'm new to calculus so sorry if my working out is stupidly wrong.)
I started simple with the equation $f(x) = \Sigma^x_{i=0} (i^2)$.
I then tried to calculate $\frac{\delta f}{\delta x}$ while $x = 2$.
I wrote down the equation like this: $\frac{(0^2 + 1^2 + (2^2).h) - (0^2 + 1^2)}{h}$. (Here I basically wrote down what $\Sigma$ is and $(2^2).h$ because $\Sigma^{x+h}_{i=0}$ would be that extra $i$ but only a small amount, right?)
I then simplified and calculated the derivatives of $i^2$ which came out as
$\frac{(2(0) + 2(1) + 2(2).h) - (2)}{h}$
$= \frac{(2 + (4).h) - (2)}{h}$
$= 2 + 4 - 2 = 4$
But this seems wrong if the derivative of sigma is the sum of its derivatives $f'(x) = \Sigma^x_{i=0} (2i)$, which would then come out as $2$ and not $4$ if $x=2$.
Again I’m sorry if the way I tried to work this out is weird, but any help with this would be appreciated; thanks.
Assuming that your function is actually defined by $$ f(x) := \sum_{i=1}^{\lfloor x \rfloor} i^{2} $$ where $\lfloor x \rfloor$ denotes the biggest integer smaller than $x$. Note that this function is not continuous at every $x\in \mathbb{N}$. Therefore calculating the derivate in these points is pointless. For every $x \notin \mathbb{N}$ you can calculate the derivative by definition $$ \frac{\mathrm{d}}{\mathrm{d}x}f(x) = \lim_{h\to 0} \frac{f(x+h)- f(x)}{h}. $$ If $h$ is small enough then you certainly have $\lfloor x+h \rfloor = \lfloor x \rfloor $, which implies $f(x+h) = f(x)$. Hence, $$ \frac{\mathrm{d}}{\mathrm{d}x}f(x) = \lim_{h\to 0} \frac{f(x+h)- f(x)}{h} = \lim_{h\to 0} \frac{0}{h} = 0. $$ Applying the linearity (sum rule) for derivatives is tricky in that case, because the number of summands depend your input variable $x$. There is no justification for this rule, if you don't have a fixed number of summands.
Anyway, if you want to apply the (in that case not justified) rule "correctly", then you have to take into account that each summand doesn't depend on $x$. So the derivative with respect to $x$ of $i^2$ is $$ \frac{\mathrm{d}}{\mathrm{d}x} i^2 = 0, \quad \text{consequently} \quad\sum_{i=1}^{\lfloor x \rfloor}\frac{\mathrm{d}}{\mathrm{d}x} i^2 = 0 $$ So luckily in that case applying this (unjustified) rule leads to a correct result.