Why we can't differentiate both sides of a polynomial equation?

Question

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.

Answer 1

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.

Answer 2

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) $4=0$.

Answer 3

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 $\left\{-\frac12,1\right\}$ However, we can't differentiate functions at isolated points of their domains!

On the other hand, consider the equation $$\sin x=\cos x\tan x.$$ The functions here agree everywhere the function on the right-hand side is defined--namely, all points except the odd integer multiples of $\frac\pi2.$ We can therefore differentiate at all such points, to obtain $$\cos x=-\sin x\tan x+\cos x\sec^2 x,$$ which one can verify to be true for all such points.

Answer 4

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

Answer 5

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.