If f is undefined at x=a, can f' exist at x=a?

[docnet]

Homework Statement

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Relevant Equations

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It isn't a homework problem, but a general problem about functions and their derivatives.

I want to say if $f$ is undefined at $x=a$, then $f$ has a discontinuity at $x=a$ so $f'$ cannot exist at at $x=a$.

For example the following function and its derivative
$$f=\frac{1}{x-a}$$ $$f'=-\frac{1}{(x-a)^2}$$
are both undefined at $x=a$.side question: Is it wrong to say that there is correspondence of discontinuities between the quotients of smooth functions and their derivatives. Is this statement wrong in general?

 

[Office_Shredder]

Write down the definition of f'(a), and see if it requires being able to compute f(a).

 

[fresh_42]

Not defined does not necessarily mean a discontinuity. $x \longmapsto \sqrt{x}$ isn't defined at $x=-1$ but this is hardly a discontinuity.

If a function isn't defined at a point, then it cannot have a derivative there.

 

[docnet]

'(z)

Write down the definition of f'(a), and see if it requires being able to compute f(a).

$$f'(a)=Lim_{h\rightarrow 0} \frac{f(a+h)-f(a)}{h}$$ is undefined if $f(a)$ is undefined

 

[docnet]

Not defined does not necessarily mean a discontinuity. $x \longmapsto \sqrt{x}$ isn't defined at $x=-1$ but this is hardly a discontinuity.

If a function isn't defined at a point, then it cannot have a derivative there.

sorry i don't follow. could you please explain the first statement?

 

[FactChecker]

You should look at the exact definitions you are using for continuity and derivatives to see if they are satisfied for a function that is undefined at a single point. In my experience, the function needs to be defined and continuous in order to have a derivative there.

 

[fresh_42]

sorry i don't follow. could you please explain the first statement?

Discontinuity and not defined are different properties.

$\sqrt{-1}$ is not defined (as real function), but you cannot (or should not) say $\sqrt{x}$ is discontinuous at $x=-1.$

\begin{align*}
f(x)=\begin{cases} 1 &\text{ if } x \leq 1\\0&\text{ if } x>1 \end{cases}
\end{align*}
is discontinuous at $x=1$ but it is defined.

 

[fresh_42]

My favorite notation of a derivative is
$$ f(x_{0}+v)=f(x_{0})+D_{x_0}(v)+r(v) \Longleftrightarrow D_{x_0}(v)=f(x_0+v)-f(x_0)-r(v)$$
You can see immediately that the right hand side needs three defined terms.

 

[docnet]

My favorite notation of a derivative is
$$ f(x_{0}+v)=f(x_{0})+D_{x_0}(v)+r(v) \Longleftrightarrow D_{x_0}(v)=f(x_0+v)-f(x_0)-r(v)$$
You can see immediately that the right hand side needs three defined terms.

to be honest I'm still confused. how did you derive this and what does $r$ and $v$ mean?

 

[fresh_42]

to be honest I'm still confused. how did you derive this and what does $r$ and $v$ mean?

Wasn't me, was Weierstraß.
\begin{align*}
D_{x_0}=f'(x_0)&=\lim_{h \to 0} \dfrac{f(x_0+h)-f(x_0)}{h}\\
D_{x_0} &= \dfrac{f(x_0+h)-f(x_0)}{h} - \underbrace{\varepsilon (h)}_{\longrightarrow 0}\\[10pt]
D_{x_0}\cdot h &=f(x_0+h)-f(x_0)- \underbrace{\varepsilon (h) \cdot h}_{=:r(h)}\\[10pt]
f(x_0+h)&=f(x_0)+D_{x_0}h+r(h)=f(x_0)+f'(x_0)\cdot h+r(h)
\end{align*}
It reads that the approximation of $f$ near $x_0,$ i.e. the value $f(x_0+h)$ where $h$ is small, equals the linear term $D_{x_0}=f'(x_0)$ in direction $h$ plus a remainder function $r(h)$. This remainder swallows the limit. It goes faster to zero than $h$ does, because
$$
\lim_{h\to 0} \dfrac{r(h)}{h}=\lim_{h\to 0} \varepsilon (h)=0
$$
The advantage of Weierstraß's notation is, that it works in all dimensions and for complex numbers, too. It also shows that a derivative is always a directional derivative. Ok, for real-valued functions in one real variable there is only one direction (or two if you distinguish plus and minus), but this defining equation remains the same if we have more than one variable. All that changes is that we write $v$ instead of $h$ for the direction, and have to divide $r(v)$ by $\|v\|$ instead.