I was working on a question which asked whether $f(x).g(x)$ is differentiable at $x=a$, given that $g(x)$ is not differentiable at $x=a$. My solution is as following:

I concluded that f(x).g(x) is not differentiable at a

I deduced that $f(x).g(x)$ is not differentiable at $a$.

But then I checked my solution using an example, and it seems that my solution was wrong.

enter image description here

In the above example, $f(x)=x-1$ and $g(x)= (mod(x-1)+1)$. Although $g(x)$ is not differentiable at $x=1$, $f(x).g(x)$ IS differentiable at $x=1$.

enter image description here

And this is how I am arriving at the conclusion that zero multiplied by something undefined is zero (at least in this case).

So my questions are:

  1. Can zero multiplied by something undefined be zero?
  2. If the answer to the above question is no, then what is the mistake in my argument?
  3. Is $f(x).g(x)$ differentiable at $x=a$, when $g(x)$ is not differentiable at $x=a$? What are the necessary conditions for $f(x).g(x)$ to be differentiable at x=a?

Note: $f(x)$ is supposed to be differentiable at x=a.

PS: I have just begun studying calculus, and I am not really a genius in Maths. So please excuse me for any grave msitakes that I may have committed above. Thanks.

  • 1
    $\begingroup$ To show why the argument does not work. Take $f(x)=x^2$ and $g(x)=\frac{1}{x^3}$. Then, the limit of $f(x)\cdot g(x)$ for $x \rightarrow 0$ does not exist although $f(x)$ tends to $0$. $\endgroup$ – Peter Jun 7 '18 at 10:36
  • 3
    $\begingroup$ $3.$ can be true , for example take $f(x)=g(x)=|x|$. Neither $f$ nor $g$ is differntiable at $x_0=0$, however the product $f(x)\cdot g(x)=x^2$ is. $\endgroup$ – Peter Jun 7 '18 at 10:43
  • $\begingroup$ most trivial example is $f(x)=0$ multiplied by any non differentiable function, but this barely counts $\endgroup$ – user438666 Jun 7 '18 at 13:16
  • $\begingroup$ @Peter, you did not tell me the mistake in my argument. You just gave an example where my conclusion doe not hold true. And regarding the third point, I wanted to know that what conditions are required for $f(x).g(x)$ to be differentiable at a point when $f(x)$ is differentiable at the point, but g(x) is not. $\endgroup$ – Aumkaar Pranav Shukla Jun 7 '18 at 13:48
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    $\begingroup$ my point was that if you have $g(x)$ equal to any non differentiable function and $f(x)=0$, $f(x)g(x)=0$ which is trivially differentiable. It's just a trivial counterexample. Your argument doesn't work because $(fg)'=f'g+fg'$ holds only if $f$ and $g$ are differentiable. One reason your question might have been downvoted is that you uploaded images instead of writing the equation, which is against policy as it renders the question harder to indicize and search. When you're trying to disprove that something is true in all cases the easiest way is often to find a counterexample $\endgroup$ – user438666 Jun 7 '18 at 15:26

There is a key issue in your post: you can't use the product rule the way you're using it. The product rule is correctly states as follows:

Let $f,g:\mathbb{R} \rightarrow \mathbb{R}$ be functions that are both differentiable at $a \in \mathbb{R}$. Then the function $h(x) = f(x)g(x)$ is differentiable at $a$ and this derivative is given by: $$h'(a) = f'(a)g(a) + g'(a)f(a)$$ where $f'$ and $g'$ are the derivatives of $f$ and $g$ respectively.

Now it may be that even in the case where $g$ is not differentiable at $a$, the function $h$ is differentiable at $a$. However the derivative cannot properly be written using the product rule.

This observation is the root of your problem. In general, you can't define quantities like $0 * [\text{undefined}]$. That's exactly what it means to be undefined! You can't do arithmetic with [undefined] in the traditional sense!

So, to go through your questions:

  1. You can't define such a quantity in general. If you're doing arithmetic with [undefined]s, you've probably made a mistake earlier in your thought process
  2. The error in your reasoning is applying the product rule in a situation where the necessary conditions to apply the product rule haven't been met
  3. In general, there isn't going to be a nice set of necessary and sufficient conditions to show when $f(x)g(x)$ is differentiable at a point in terms of the differentiability of $f(x)$ and $g(x)$. See the following pathology:

\begin{align*} f(x) &= \begin{cases} 0 & \quad \text{if $x$ is rational} \\ 1 & \quad \text{if $x$ is irrational} \end{cases} \\ g(x) &= 1 - f(x) \end{align*} Notice that both $f$ and $g$ are nowhere differentiable, but we have $h(x) = f(x)g(x) = 0$ for all $x$. $h$ is differentiable everywhere.

  • $\begingroup$ I have noticed that if f(x) is differentiable at x=a, and g(x) is not; then f(x).g(x) is differentiable at x=a whenever f(a)=0. Can you please give an example where f(x).g(x) is differentiable at x=a when f(a) is not 0? $\endgroup$ – Aumkaar Pranav Shukla Jun 8 '18 at 15:39
  1. Zero multiplied by something undefined is undefined. So the answer is no.

  2. Your mistake lies in a wrong application of the theorem which says: If $h=fg$ and $g$ and $f$ are differentiable at $a$, then $h$ is differentiable at $a$ and $h'(a)=g'(a)f(a)+g(a)f'(a)$. In your case, the hypotheses of the theorem are not fulfilled.

  3. Depends on the situation. You show one example in which none are differentiable or continuous and the product is everywhere differentiable. What is not possible is that $f$ and $g$ are differentiable and $h$ is not. The product rule theorem gives you sufficient conditions, not necessary. Both differentiable, then the product is also. If any one is not differentiable, $h$ could either be or not.

  • $\begingroup$ Thanks! Can you please check the update I have made in point 3, and then edit your answer accordingly? $\endgroup$ – Aumkaar Pranav Shukla Jun 7 '18 at 14:40
  1. Yes, you showed it yourself! Another fun example would be to look at $$ x\cdot\sin(1/x) $$ at $x=0$.
  2. Your mistake lies in the fact that you have just written down $f(a) g'(a)$. Now, this does not exist but it might be the case that $$ \lim_{x\to a} f(x) g'(x) $$ exists.
  3. Really depends on the situation, but if $f(a)$ is non-zero, it will definitely be not the case. In the other case, you can find examples at both ends.

I hope this helps!

  • $\begingroup$ Your answer differs from the other answer that I have got. Which one is right? $\endgroup$ – Aumkaar Pranav Shukla Jun 7 '18 at 13:53
  • $\begingroup$ $\lim_{x\to 0} f(x)g(x)$ is not the same thing as $f(0)*g(0)$. $\lim_{x\to 0} x\sin\frac 1x = 0$ but $0*\sin \frac 10$ is undefined. (Also undefined is not the same thing as infinitely large). An easier example is simple $x^2*\frac 1x$. $\lim_{x\to 0}x^2*\frac 1x =0$ but $0^2*\frac 10$ is undefined. $\endgroup$ – fleablood Jun 7 '18 at 15:51
  • $\begingroup$ @Aumkaar Pranav Shukla I am pretty sure I answered your question correctly, at least I addressed the exact parts you asked. Joe also provides a good explanation of the stuff that is going on here. Although he answers your question 1 wrongly. He says: in general, this is undefined. While you ask whether there is a situation in which it is defined and to that question, the answer is a simple yes, it is possible. $\endgroup$ – Stan Tendijck Jun 7 '18 at 16:54
  • $\begingroup$ I think that the product rule can be applied if both limits exist (instead of filling in. $a$, you could view it as a limit for $x\to a$. So, that's also a point I would like to address $\endgroup$ – Stan Tendijck Jun 7 '18 at 16:57
  • $\begingroup$ Strictly speaking, you cannot write $f(a)\cdot g'(a)$ if the latter does not exist. However, you can make sense of this expression to interpret it as a the limit for $x\to a$ of $f(x) g'(x)$. That's what I am trying to say essentially $\endgroup$ – Stan Tendijck Jun 7 '18 at 17:05

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