Equivalence of an alternative definition of the derivative.

Question

My calculus professor defined the derivative of a real function $f$ as follows :

“Let $f\colon \mathbb{R} \longrightarrow \mathbb R$ be a function. Then $f$ is said to be differentiable at a point $a \in\mathbb R$ if $\exists$ a function $\psi$ such that :

1) $(\forall h \in\mathbb{R}): f(a+h) = f(a) + mh + hΨ(h)$

2) $\psi$ is continuous at $0$

3) $\psi(0) = 0$

Then we denote the derivative of $f(x)$ as $f'(x) = m$”

My problem in this definition is as follows: I can see that by rearranging the terms $f(a)$ and then dividing by h we get the regular definition of the limit. I do not see what is the significance of $\psi$.

Answer 1

From the new definition,

$$\psi(h)=\frac{f(a+h)-f(a)}h-m.$$

If you take the limit for $h\to0$,

$$\lim_{h\to0}\psi(h)=\lim_{h\to0}\frac{f(a+h)-f(a)}h-m=f'(a)-m.$$

So $m$ plays the role of the derivative and $\psi$ is an error term, i.e. the adjustment required when $h$ is nonzero. For $m$ to equal the derivative, we need to error term to vanish for $h\to0$.

For the sake of illustration, a sinusoid, the tangent at $x=1$ and the error function.

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Technical note:

The properties 2) and 3) of $\psi$ are another way to express that $\lim_{h\to0}\psi(h)=0$.

Answer 2

The fact that $\psi$ is continuous at $0$ is useful to prove theorems. Try to prove the chain rule using the standard definition. And then try to prove it using the new definition. You will see that it is much easier to get a proof using the new definition.