Is this a valid definition of a derivative

algebra-precalculuscalculus

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.

Best Answer

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 $f'(0)$.

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