Approximation of a non differentiable function

Question

Can you please tell me if/where I am wrong about this? Let $f: \mathbb{R}^2\to \mathbb{R}$ a function. Let $f$ be derivable at $\left( \frac{6}{5}, \frac{6}{5}\right)$ but not differentiable at that point.

Which, among $f\left(\frac{6}{5}, \frac{6}{5}\right)$ and $f\left(\frac{6}{5}, 1\right)$ can be approximated in a good way, linearly, by the expression $$f(x, y) \approx f\left(\frac{6}{5}, \frac{6}{5}\right) + h\left(x - \frac{6}{5}\right) + k \left(y - \frac{6}{5}\right)$$ and for which values of $h, k$? Why is the approximation good (IF it is good)?

Attempts

So, in our course the professor used the words derivable and differentiable. She said "derivable" means the function admits partial derivatives at all points and directions, whereas "differentiable" has the "usual" meaning in $\mathbb{R}^n$ (the one we all learnt: there does exist a linear map such that....).

So, this being said what I would say is that only $f\left(\frac{6}{5}, 1\right)$ can be approximated in a linear way because if by hypothesis $f$ is not differentiable at $\left(\frac{6}{5}, \frac{6}{5}\right)$, then there do not exist the tangent planes at that point, hence there is no linear approximation of $f(x, y)$ at that point.

Since $f$ is derivable, then the partial derivatives at $\left(\frac{6}{5}, \frac{6}{5}\right)$ exist in every direction, hence I would say $h = \frac{\partial f}{\partial x}\left(\frac{6}{5}, \frac{6}{5}\right)$ and $k = \frac{\partial f}{\partial y}\left(\frac{6}{5}, \frac{6}{5}\right)$.

Yet I don't know if I thought well or not... Also, I don't understand the question "why is the approximation good"?

Thank you as always...

Yes, I did, and I cannot understand the difference between "differentiable" and "derivable" — regardless of "what the teacher said". — Dan Asimov, Apr 12 '24 at 20:12
@DanAsimov I think it's obvious. Derivable = partial derivatives exist at all points. This doesn't imply it's also differentiable. — Heidegger, Apr 12 '24 at 20:13
I have no idea what you mean. (And I have taught calculus and advanced calculus many, many times.) Can you offer one example of a function (with domain and codomain stated) that is one but not the other? — Dan Asimov, Apr 12 '24 at 20:14
@DanAsimov I don't know what to say. Ignore the term then. Just take the fact that partial derivatives exist but it's not differentiable. — Heidegger, Apr 12 '24 at 20:17
@DanAsimov
By differentiability theorem if partial derivatives exist and are continuous in a neighborhood of the point then (i.e. sufficient condition) the function is differentiable at that point.

The existence of partial derivatives doesn't suffice.

Take this example — Heidegger, Apr 12 '24 at 20:18
@DanAsimov
$$
f(x,y)= \begin{cases} \dfrac{2x^2y+y^3}{x^2+y^2} & \text{if $(x,y) \neq (0,0)$}\ 0 & \text{if $(x,y) = (0,0)$}\ \end{cases} $$ — Heidegger, Apr 12 '24 at 20:18
Heidegger, here is the problem: In mathematics, if all partial derivatives exist (at all points of the domain), then a function is differentiable. So I am afraid that I don't know what you are proposing. — Dan Asimov, Apr 12 '24 at 20:19
@DanAsimov Take the previous example: not only partial derivatives exist, but at $(0, 0)$ the directional derivaties ALL exist. Yet it's not differentiable. Are you sure you taught "advanced" calculus? — Heidegger, Apr 12 '24 at 20:20
@DanAsimov you're either being incredibly obtuse or you would benefit from sitting in an introductory calculus lecture once again — Jose Avilez, Apr 12 '24 at 20:31
https://math.stackexchange.com/questions/3831023/can-a-function-have-partial-derivatives-be-continuous-but-not-be-differentiable — Jose Avilez, Apr 12 '24 at 20:32
@JoseAvilez Do you think my reasoning in the question I posted is then correct? — Heidegger, Apr 12 '24 at 20:50
The answer to the actual question is this: Values $(h,k)$ as you said except with one equal to zero. In other words, you’re restricting to lines through the point parallel to the axes and using the linear approximation of a single-variable function. — Ted Shifrin, Apr 12 '24 at 23:54
@TedShifrin Can I ask just a clarification? Why do I have to take one of them equal to zero? I didn't understand why I cannot use them both, since I'm, even if slightly, away from the point of non differentiability — Heidegger, Apr 14 '24 at 10:56
Btw, "derivable" (as opposed to "differentiable") is not very standard, so it's reasonable for people to ask what you mean by it, I think. — paul garrett, Apr 14 '24 at 20:07

score 1 · Answer 1 · answered Apr 14 '24 at 19:51

If the function $f$ has partial derivatives at $(a,b)$, then the function $g(x)=f(x,b)$ is differentiable at $x=a$ and the function $h(y)=f(a,y)$ is differentiable at $y=b$. Thus, restricting to these lines, you have the usual good linear approximation coming from the single-variable derivative.

However, if the function $f$ fails to be differentiable at $(a,b)$, it may not even be continuous there. The classic example is the function $$f(x,y) = \begin{cases} \frac{xy}{x^2+y^2}, & (x,y)\ne (0,0) \\ 0, & (x,y)=(0,0)\end{cases}.$$ Both partial derivatives at the origin are $0$, but this function is not even continuous at the origin: For example, on the line $y=mx$, the function is constant with a jump at the origin: $$f(x,mx) = \begin{cases} \frac m{1+m^2}, & x\ne 0 \\ 0, & x=0\end{cases}.$$ But the partial derivatives would give you a linear approximation by the $0$ function; is that a good approximation even the slightest distance away from the origin? You can explore this further with many other examples; try an example of a function that is continuous but not differentiable at the origin. You can find literally dozens of posts on MSE discussing this phenomenon.

Approximation of a non differentiable function

1 Answers1