# Can a function have a derivative where it has no value?

Consider the following function: $$f(x) = x^2 \mid x \in \mathbb{R}, \ x \ne 0$$

The derivative at $x=0$ seems to want to be zero, in the same way that $\lim \limits_{x \to 0} f(x) = 0$

However, when I look at the definition of the derivative, this doesn't seem to work: $$f'(x) = \lim \limits_{\Delta x \to 0} \frac{f(x+\Delta x )-f(x)}{\Delta x}.$$ The function isn't defined at $f(x)$, so $f'(x)$ is also undefined. Would it make any sense to replace the $f(x)$ in the definition with $\lim \limits_{x \to 0} f(x) = 0$? Then, I suppose we'd have $\lim \limits_{x \to 0} f'(x) = 0$?

Would it be permissible? Would there be any point?

• I've arbitrarily defined it with a hole at $x=0$. Did I not represent that correctly? Apr 22, 2018 at 6:20
• What if I set f(0) = -1? Just to be pesky? The limit is still zero but it seems like the derivative is a mess. Apr 22, 2018 at 6:22
• If you set $f(0)=-1$, then the new function $f$ is not continuous at $x=0$, hence not differentiable at $x=0$. But take note of MPW's answer. If you extend the function $f$ by defining it at $x=0$, ir's no longer $f$, so you should use a new letter, so as not to cause confusion. Apr 22, 2018 at 6:25
• A naive approach might instead be to consider $$\lim_{h\to 0} \frac{f(x +h)-f(x-h)}{2h}.$$ Apr 22, 2018 at 6:58
• @Shalop Your approach is not equivalent to the commonly accepted definition. Compare the results for the modulus function: $\mathbb R\ni x\mapsto|x|\in\mathbb R$. Your expression evaluates to $0$ while the original is 'undefined'. Apr 23, 2018 at 8:17

If a function $f$ is not defined at a point $x_0$ we can't evaluate the derivative at $x_0$ because in the definition of the derivative at $x_0$ we need $f(x_0)$.

However if $f$ is continuous in $0<|x-x_0|<r$ with $r>0$ and the limit $\lim_{x\to x_0}f(x)=L$ exists we can extend $f$ to a continuous function in $|x-x_0|<r$ by letting $f(x_0):=L$.

Moreover if $f$ is also differentiable in $0<|x-x_0|<r$ and the limit $\lim_{x\to x_0}f'(x)=a$ then the extended function is also differentiable at $x_0$ and its derivative at $x_0$ is equal to $a$.

No.

You can certainly extend $f$ to a function $g$ as you indicate, but then you are really computing $g'(0)$, not $f'(0)$.

In that case, $f'$ agrees with $g'$ on their common domain, which excludes $x=0$.

Remember, if a function is differentiable at $x_0$, then it is continuous at $x_0$. In particular, it is defined at $x_0$.

As has already been said, and as you have found yourself, the ordinary derivative is not defined. However, the symmetric derivative is defined in your case: $$f'_{\text{symmetric}}(x_0) := \lim_{h\to 0} \frac{f(x_0+h)-f(x_0-h)}{2h}$$

A more general derivative is also defined (I'll call it excluding; I don't know if it has a name): $$f'_{\text{excluding}}(x_0) := \lim_{h,k\to 0\\h,k \neq 0} \frac{f(x_0+h)-f(x_0+k)}{h-k} = \lim_{x_1,x_2\to x_0\\x_1,x_2 \neq x_0} \frac{f(x_1)-f(x_2)}{x_1-x_2}$$

• I don't know if it has a name --- From my answer to “Strong” derivative of a monotone function: "Besides strong derivative, three other names I’ve encountered in the literature are unstraddled derivative (Andrew M. Bruckner), strict derivative (Ludek Zajicek and several people who work in the area of higher and infinite dimensional convex analysis and nonlinear analysis), and sharp derivative (Brian S. Thomson and Vasile Ene)." Apr 22, 2018 at 8:19
• I don't think it's necessary to specify $h,k\neq 0$ in the limit subscript, is it? I would expect it to be implicit in the definition of a limit that the changing variable doesn't actually take on the value which it is approaching. Apr 22, 2018 at 8:38
• @DavidZ $\lim_{x_1,x_2 \rightarrow x_0}$ isn't quite standard notation. So it might be implying you get equal and defined results from $\lim_{x_1 \rightarrow x_0} \lim_{x_2 \rightarrow x_0}$ and $\lim_{x_2 \rightarrow x_0} \lim_{x_1 \rightarrow x_0}$, but that's not in general true. Or more likely it would mean $\lim_{(x_1, x_2) \rightarrow (x_0, x_0)}$, but this allows approach from any path on $\mathbb{R}^2$, including straight horizontal and vertical. Apr 22, 2018 at 10:52
• It seems to me that the defintion of $f'_{\text{symmetric}}$ says that if $f(x) = \lvert x\rvert,$ then $f'_{\text{symmetric}}(0)=0.$ But the definition of $f'_{\text{excluding}}$ then would say the derivative doesn't exist at $x=0,$ which seems a more sensible answer. Apr 22, 2018 at 13:14
• Thanks @aschepler for giving a good answer to why I wrote $h,k \neq 0$. Apr 22, 2018 at 16:43

You function does have a derivative, just not a "classical" or "strong" one. Instead your function has what's called a "weak derivative", which is technically not a single function but rather a class of functions for technical reasons from measure theory about zero-measure sets, but this class usually (in practical problems) has a single continuous or piecewise continuous representative. In your case, the function $$g(x) = 2x$$ is the continuous representative of the class of weak derivatives of your function $$f$$, namely because for any "test function" $$\phi$$ (i.e. a smooth and compactly supported function) on $$\mathbb{R}$$ we see that the relation of integration by parts holds:

$$\int_\mathbb{R} f(x) * \phi'(x) dx = \int_\mathbb{R} x^2 * \phi'(x) dx = - \int_\mathbb{R} 2x * \phi(x) dx = \int_\mathbb{R} g(x) * \phi(x) dx$$