# Spot the mistake

**Calculus**Level 1

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

\[\begin{align} 1^2&=1\\ 2^2&=2+2&\text{(2 times)}\\ 3^2&=3+3+3&\text{(3 times)} \end{align}\]

**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

\[ 2 = 1. \]