Spot the mistake

Here's my attempt at proving that $2=1$. In which step did I first commit a flaw in my logic?

$\begin{aligned} 1^2&=1\\ 2^2&=2+2&\text{(2 times)}\\ 3^2&=3+3+3&\text{(3 times)} \end{aligned}$

Step 1: For any positive integer:

$x^2=x+x+\cdots+x\quad\quad\text{(x times)}$

Step 2: Now, differentiating with respect to $x$:

$2x=1+1+\cdots+1\text{(x times)}.$

Step 3: Summing this up, we get

$2x = x.$

Step 4: Dividing by $x$, we get

$2 = 1.$