Solving $\frac{df}{dx} = 2x$ using the definition of a derivative.

Question

Although it's pretty straightforward to solve the above equation but I was wondering, how about if we use the limit definition of a derivative and then try to solve it. So, the equation looks like $$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h} = 2x$$ Now how should I go about solving it? IMO guessing might be one of the ways but I am not aware of any of the "formal" ones. Any kind of help will be appreciated.

Answer 1

First note that $\frac{d}{dx}x^2=2x$; this can be established from the definition of derivative.

Now we have that $\frac{d}{dx}[f(x)-x^2]=0$, since the linearity of differentiation can be established from the definition of derivative.

Finally we need to prove from the definition of derivative, and the other properties of real numbers this version of the Mean Value Theorem:

Suppose that $g(x)$ is differentiable, and that $g'(x)=0$ for all $x$. Then for any $a,b$ we have $g(a)=g(b)$.

We then have that $f(x)-x^2=k$ for some constant $k$.

I do not believe that there is any proof that does not use the key theorem of the differentiable calculus, namely the Mean Value Theorem.

Answer 2

Take $f(x)=ax^2+bx+c$ because of $f'=2x^1$ so $$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h} = \\\lim_{h\to 0}\frac{a(x+h)^2+b(x+h)+c-(ax^2+bx+c)}{h} =\\ \lim_{h\to 0}\frac{2axh+ah^2+bh}{h} =\\\lim_{h\to 0}2ax+b+ah =2x\\a=1\\b=0\\c \in \mathbb{R}$$

Answer 3

This sounds like a fun exercise $$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h} = 2x$$

let’s assume $f(x)$ is a polynomial

so $f(x) = ax^n$

$$\lim_{h\to 0}\frac{a(x+h)^n-ax^n}{h} = 2x$$

we can write it like thing where all higher terms of h is removed because they will go to zero

$$\lim_{h\to 0}\frac{(ax^n + ax^{n-1}h+…)-ax^n}{h} = 2x$$

$$\lim_{h\to 0}\frac{anx^{n-1}h+…}{h} = 2x$$

$$\lim_{h\to 0} anx^{n-1} + … = 2x$$

$$ anx^{n-1} = 2x $$

now we can solve for $f(x)$, $n-1 = 1$, $a = 1$, $f(x) = x^2$