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Usually in most introductory Calculus courses, a definition of differentiability at a point $a$ is defined, as follows :

A function $f$ is differentiable at $a$ if $f'(a)$ exists

As a corollary of this :

A function $f$ is not differentiable at $a$ if $f'(a)$ does not exist


However can differentiability be defined another way, in terms of continuity of the derivative function?

Possible Alternate Definition

A function $f$ is differentiable at $a$ if $f'(x)$ is continuous at $a$

Stated more formally, given $f : \mathbb{R} \to \mathbb{R}$, $f$ is differentiable $\text{iff}$

$$\lim_{x\to a} f'(x) = f'(a)$$


A simple example to illustrate this would be to look at $f(x) = |x|$.

$$f'(x) =\begin{cases} \ \ \ 1 &\text{if} & x > 0 \\ -1 &\text{if} & x < 0 \\ \end{cases}$$

And using the alternate definition, we can show that since the limit doesn't exist $$\lim_{x \to 0} f'(x) \not= f'(0)$$ and thus $f'(x)$ is discontinous at $x=0$, and hence not differentiable at $x=0$.


Note: I haven't looked up whether this is actually a definition or not (this is just something I thought up), as far as I know, differentiability in most textbooks are not defined in this way. If this possible alternate definition, that I've given is incorrect, in any way whatsoever, or if it doesn't generalize well (conceptually that is) outside of single-variable calculus for real-valued functions, please feel free to tear it apart completely.

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    $\begingroup$ There exists an everywhere differentiable function whose derivative is discontinuous. Your notion is almost the same as what mathematians call $C^1$ functions (continuously differentiable functions). $\endgroup$ – Sangchul Lee May 20 '16 at 22:53
  • $\begingroup$ So you want to say the existence of $f'(x)$ is different from saying $f$ is differentiable at $x?$ I fail to see the point of this. $\endgroup$ – zhw. May 20 '16 at 23:55
  • $\begingroup$ @Perturbative: Wait, did you just wrote down that $f'(0)=1$ when $f(x)=|x|$? $\endgroup$ – Mathematician 42 May 20 '16 at 23:58
  • $\begingroup$ @Mathematician42, My bad, made an error when I was typing the post out. It has been fixed now. $\endgroup$ – Perturbative May 21 '16 at 0:09
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There are functions which are differentiable at a point $a$ but the derivative $f'(x)$ is not continuous at $a$. For instance, $f(x) = x^2 \cos(1/x)$ when $x\neq0$, and $f(0) = 0$. Using the definition: $$ f'(0) = \lim_{h\to 0} \frac{f(h) - f(0)}{h} = \lim_{h\to 0}\frac{h^2\cos(1/h)}{h} = \lim_{h\to 0}h \cos(1/h) = 0 $$ (the limit is zero because the $\cos(1/h)$ term is bounded by $1$ in absolute value). However, away from zero, $$ f'(x) = 2x \cos(1/x) - \sin(1/x) $$ As $x\to 0$, the first term tends to zero but the second has no limit. Therefore $\lim_{x\to 0} f'(x)$ does not exist.

However, if both $\lim_{x\to a^+} f'(x)$ and $f'(a)$ exist, they must agree. Proof: For any $x$ sufficiently near $a$ but greater than $a$, the difference quotient $\frac{f(x)-f(a)}{x-a}$ is equal to $f'(c_x)$ for some point $c_x$ in $(a,x)$ (Mean Value Theorem). As $x\to a^+$, $c_x \to a$, so $f'(a) = \lim_{x\to a} f'(x)$.

Similarly, if $\lim_{x\to a^-} f'(x)$ and $f'(a)$ exist, they must agree. The proof is identical, mutatis mutandis.

This means that if $\lim_{x\to a^+} f'(x)$ and $\lim_{x\to a^-} f'(x)$ exist, but disagree, then $f$ is not differentiable at $a$. For it it were, $f'(a)$ would need to be equal to two unequal things.

Therefore, your proof that $|x|$ is not differentiable at $0$ is legitimate, but it relies on the Mean Value Theorem rather than the definition.

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