Basically, an integral adds up infinitely small pieces, whereas a sum adds up distinct pieces.
$$
\int_1^{\infty} xdx\,
$$
Will add up all of the area under f(x)=x, whereas
$$
\sum_{x=1}^{\infty} x\,
$$
Will add up each value for x from whatever you start, until you stop.
For $f(x)=1/x$ this is the difference between the area under 1/x, and the series $1 + 1/2 + 1/3 + 1/4 +...+ 1/n $ (the harmonic series).
Both of these values diverge.
The issue here is that the derivative $y'(1)$ is not defined. I'll start with the definition and then discuss why the differentiation rules don't apply at $x = 1$. First, for the derivative to exist at $x = 1$, then by definition,the following limit has to exist:
$$\lim_{h\to 0}\frac{y(1+h) - y(1)}{h} \tag{1}.$$
Now, in order for $(1)$ to exist the left and right limits have to exist and agree. So, we can look at the left and right derivatives:
Right Derivative:
$$y'_{+}(1) = \lim_{h\to 0^{+}}\frac{f(1+h) - f(1)}{h} = \lim_{h\to 0^{+}}\frac{2(1+h) - 2}{h} = \lim_{h\to 0^{+}}\frac{2h}{h} = 2$$
There's no trouble there, however we run into problems with the left derivative:
Left Derivative:
\begin{align}
y'_{-}(1)&=\lim_{h\to 0^{-}}\frac{y(1+h) - y(1)}{h}\\
&= \lim_{h\to 0^{-}}\frac{(1+h)^2 - 2}{h}\\
&= \lim_{h\to 0^{-}}\frac{1 +2h+h^2-2}{h}\\
&= \lim_{h\to 0^{-}}\frac{h^2+2h-1}{h}\\
&= \lim_{h\to 0^{-}}\left(h + 2 - \frac{1}{h}\right)\\
&= \infty
\end{align}
So, since the left and right derivatives don't agree, we can say that $y$ is not differentiable at $y = 1$. As has been pointed out in the comments we could also have concluded that from continuity, because if $y'(1)$ had existed, then $y$ would need to be continuous at $x = 1$, and because it is not we can immediately rule out differentiability there.
Now, to understand why it doesn't work to take the derivatives of $2x$ and $x^2$ and take a limit, we start by noting that we can certainly write
$$y'(x) = \begin{cases}
2x,&-\infty<x<1\\
2, &1<x<\infty
\end{cases}$$
However, if we were to write $$y'(1) = \lim_{x\to 1}y'(x)$$ we would be making two unfounded assumptions:
First, that $y$ is actually differentiable at $x = 1$, and so writing $y'(1)$ makes sense, and second that $y'$ is continuous. We don't know a priori that either of those things are true, and in fact as previously mentioned, we know that $y'(1)$ can't exist because $y$ isn't continuous at $x = 1$.
To give a bit more of an intuitive explanation, remember that the derivative at a point relies on the value of the function at that point. So, when we use $2x$ for the derivative on the left, instead of using $y(1) = 2$, which lies on the right piece, we're really using $\lim_{x\to 1^{-}}y(x) = 1$ on the left piece.
Best Answer
Ok, so the intuitive pictorial idea of a derivative of a function $f$ at a point $x$ is taking the slope of the graph of $f$ at the point $x$. Now this is fine for guiding intuition but it isn't mathematically rigorous so let's try to be more rigorous.
So firstly what we do is we try to approximate the slope at a point, you may have heard of secant lines? That is what we are doing here. So to approximate the slope we take a small value $h$ and we take the slope between $f(x)$ and $f(x+h)$. Remember that gradient is "rise over run?" so our approximation is our rise $f(x+h)-f(x)$ over our run $(x+h)-x=h$. Hence our approximation is: $$ \frac{f(x+h)-f(x)}{h} $$ I hope you recognize this equation.
Ok so we have an approximation, now the idea is that if our function $f$ is smooth enough at $x$ then as $h$ gets smaller, this approximation is going to get better. If this is the case wouldn't we want to take $h=0$ to get the best approximation? Unfortunately we have a problem here, if you substitute $h=0$ into the approximation above we get $\frac{0}{0}$ which is bad. So what are we going to do?
Well what we do is hope that as $h$ gets close to 0, that our approximation starts to settle down and get close to the gradient we are trying to approximate. To this end we ask does $$ \lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h} $$ Exist? If it does, then our approximation is good, and we say that $f$ is differentiable at $x$. And we call the value $ f'(x)=\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h} $ the derivative of $f$ at $x$.
If the limit does not exist then it means that our function is not smooth enough to take good approximations of the gradient, and if we can't approximate it well we say it doesn't exist, and we say $f$ is not differentiable at $x$.
So a function is differentiable if the limit above exists, and the derivative is the value of the limit.
This clear things up?