# Is this a valid definition of a derivative?

I was messing around with differentiation with first principles and I thought of a 'new' way to define the derivative.

Consider a point $x$. Then instead of small change $h$, let it be a scale factor $h$. Then a derivative can be defined from the following:

$$f'(x) = \lim_{h \to 1} \frac{f(hx)-f(x)}{hx-x}$$

I tried letting $f(x)= x^2$ and it seemed to work. How valid is this new form of First Principles? Is this the same as taking the derivative form First Principles? Any ideas or answers would be appreciated.

• It's a bit useless when $x=0$.... – Lord Shark the Unknown Sep 1 '18 at 15:22

We can make a change of variable:

• $\epsilon = hx - x$
• $h = 1 + \frac{\epsilon}{x}$

The change of variable transformations are well-defined, continuous, and invertible for any fixed $x \neq 0$, so there are no technical issues performing the substitution if we only ask for the derivative in that case.

$$\lim_{h \to 1} \frac{f(hx)-f(x)}{hx-x} = \lim_{\epsilon \to 0} \frac{f(x + \epsilon)-f(x)}{\epsilon}$$

So your definition is indeed equivalent to the usual one! (in the $x \neq 0$ case)

Of course, your formula fails badly when considering $f'(0)$.

Applying L'Hospital to your definition we obtain $\frac{xf'(hx)}{x}$, which tends to $f'(x)$ for $h\to 1$, so your definition yields the right value. The problem is you can't differentiate any function at $x=0$ this way...

• To work-around finding $f'(0)$: could you first find $f'(x)$ then substitute $x=0$? – Landuros Sep 1 '18 at 14:55
• If $f'$ is continuous and bounded near $0$, then I believe that works, but really, you are not gaining anything. BTW the answer of @Hurkyl is better than mine, as it works when $f'$ is not continuous. – Kusma Sep 1 '18 at 14:59