Dangers of Differentiation

The following shows an interesting line of reasoning posed by Math Lazy Hazel on Twitter.

What is faulty in this demonstration that 2 = 1? I asked Gemini via this prompt:

There is a flaw in the logic of the Mathematics on this uploaded file. Can you ascertain what it is? On the surface the differentiation seems logical enough.

The response was quite well explained.

Of course. The flaw in the mathematical logic lies in the incorrect application of differentiation to a sum with a variable number of terms. Here’s a breakdown of the error: 

$\mathbf{\text{The Flaw Explained}}$ 

The core mistake happens when moving from the second to the third line of the "proof." 

$\mathbf{\text{The Setup}}$: 

The proof starts by defining $x^2$ as the sum of $x$ added to itself $x$ times: $$x^2=\underset{x\text{ times}}{\underbrace{x+x+\cdots +x}}$$ This definition itself is only truly valid for integer values of $x$. 2.

$\mathbf{\text{The Incorrect Differentiation}}$: 

The next step applies the derivative with respect to $x$ to both sides: $$\frac{d}{dx}{x}^2=\frac{d}{dx}\left(\underset{x\text{ times}}{\underbrace{x+x+\cdots +x}}\right)$$ The left side correctly becomes $2x$. However, the right side is differentiated incorrectly. The proof mistakenly treats the number of terms in the sum as a constant. The rule for the derivative of a sum, $\frac{d}{dx}\sum f_i=\sum \frac{d}{dx}{f}_i$, only works when the number of terms is constant. In this case, the number of terms is $x$, the very variable you are differentiating with respect to. You cannot simply differentiate each term individually and add them up while ignoring that the number of terms is also changing.

$\mathbf{\text{Correct Approach}}$

To correctly differentiate the expression on the right, you must use the product rule, because it's effectively a product of two functions of $x$: the value of each term ($x$) and the number of terms ($x$). Let the function be $f(x)=x\cdot x$. Using the product rule $(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }$: $$\begin{array}{rl}\frac{d}{dx}(x\cdot x)& =\left(\frac{d}{dx}x\right)\cdot x+x\cdot \left(\frac{d}{dx}x\right)\\ & =(1\cdot x)+(x\cdot 1)\\ & =x+x=2x\end{array}$$ This correct differentiation results in $2x=2x$, which is a valid identity and does not lead to the false conclusion that $2=1$. In short, the proof is flawed because it misapplies the rules of differentiation by treating a variable number of terms as a constant.