Argthtjtr

Suppose we had the equation below and we are going to differentiate it both sides:
begin{align}
&2x^2-x=1\
&4x-1=0\
&4=0
end{align}

This problem doesn't seems to happens with other equation like $ln x =1$ or $sin x = 0$, we can keep differentiating these two without getting "$4=0$", for example. This why I asked about polynomials.

PS: I'm not trying to solve any of these equations by differentiating then. But differentiation or integration helps and solving equations?

I remember that sometimes to solve trigonometry equtions like $sin x = cos x$ we had to square both side so we could use the identity $sin^2x + cos^2x =1$. Even thought squaring appears to make it worse because we have a new root.

It's important to remember that we can only differentiate functions. When you write the expression
$$
2x^2-x=1
$$
you are no longer dealing with a function. Instead, this expression describes only the solutions $x$ to a given equation. For instance,
$$
f(x) = 2x-x^2
$$
is a function, but $2x-x^2 = 0$ is not.

When two functions intersect, they don't have to have the same slopes. For example, $y=x^2, y=x$.

The kicker is that our domain of truth isn't "big enough" to allow it.

From your example, the functions on both sides only agree on $left{-frac12,1right}$ However, we can't differentiate functions at isolated points of their domains!

On the other hand, consider the equation $$sin x=cos xtan x.$$ The functions here agree everywhere the function on the right-hand side is defined--namely, all points except the odd integer multiples of $fracpi2.$ We can therefore differentiate at all such points, to obtain $$cos x=-sin xtan x+cos xsec^2 x,$$ which one can verify to be true for all such points.

In general if two functions $f$ and $g$ agree at point $a$ they need not have the same derivative there i.e. $f(a)=g(a)$ does not imply $f'(a)=g'(a)$. This is readily seen by taking $f(x)=x$ and $g$ to be the constant function at $1$. Then for example $f(1)=g(1)$ but $1=f'(1)neq g'(1)=0$.

We shouldn't be surprised since to compute the derivative of a function $f$ at a point $a$ we need to know how $f$ behaves in a neighbourhood $(a-h, a+h)$ for some $h>0$ of that point. Knowing the value of the point is not enough. There are many ways to draw a differentiable curve through a point.

In the event that the two functions do agree on a neighbourhood, then the desired claim does follow.

Differentiation isn't an algebraic operation like squaring or addition. Same thing with integration.

You found a counterexample yourself: if you could solve $2x^2-2x=1$ with diff. then you would've gotten $x=-1/2,1$, not (the contradictory statement) $4=0$.