Another $1=2$ proof

Question

So a friend shows me this :

$x^4= x^2+x^2+ \cdots +x^2 $ ( i.e. $x^2$ added $x^2$ times)

Now take the derivative of both side;

$4x^3 = 2x + 2x + \cdots + 2x $;

So $4x^3 = 2x^3 \cdots $(1)

And so dividing by $x^3$ gives $2=1 \cdots $(2).

I know we can't divide by 0 so that makes (2) false, but to show that (1) is false too ?

Answer 1

1. First of all, the statement

$x^2$ added $x^2$ times

makes sense only if $x^2$ is a positive integer. Else if $x^2$ is not a positive integer, then the statement is meaningless.

2. Moreover, from $(1)$, we have that $4x^3=2x^3 \Rightarrow 4x^3-2x^3=0 \Rightarrow 2x^3=0 \Rightarrow x^3=0$

And hence division by $x^3$ in the next step is meaningless.

Answer 2

You can only do the first step if $x^2$ is an positive integer. So since it doesn't hold for all $x$ you can't take the derivative of both sides like that. It doesn't even make sense to talk about "$x^2$ times" if $x^2$ is not a natural number. To take the derivative of both sides you'd need equality in an entire interval.