-2
$\begingroup$

If we have a function $g(x)$ defined by $g(x) = f_1(x)f_2(x)$ where $f_1(x)$ and $f_2(x)$ are non-differentiable at some points, can $g(x)$ ever be differentiable everywhere? Intuitively using product rule, derivative of $g(x)$ would be $f_1(x)f_2'(x)+f_1'(x)f_2(x)$ which means that at any points where $f'_{\text{1 or 2}}(x)$ is not differentiable, the derivative cannot exist, but I have 2 doubts about this result.

If a function $g(x)$ can be defined as $g(x) = f_1(x)f_2(x)$, does that necessarily mean the derivative can be expressed as $f_1(x)f_2'(x)+f_1'(x)f_2(x)$? I know that for differentiable functions $f'_{\text{1 and 2}}(x)$ this is true, but have no idea for non-differentiable.

Can multiplying the "invalid" $f'_{\text{1 or 2}}(x)$ by another function make it "valid"? I don't see a way to proceed further with either of these 2 problems.

New contributor
Paolo Mancini is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
$\endgroup$
6
  • $\begingroup$ Try to build an example where the bad arguments for $f_1$ are all less than $0$ and $f_1$ identically $0$ for positive arguments, and $f_2$ the reverse (identically $0$ for negative arguments and all bad arguments are positive). $\endgroup$ Commented Nov 26 at 13:36
  • 2
    $\begingroup$ $x\cdot\frac1x=1$ $\endgroup$ Commented Nov 26 at 13:36
  • $\begingroup$ @lulu what is meant by bad arguments? $\endgroup$ Commented Nov 26 at 13:42
  • $\begingroup$ @wasn'tme i should've thought about that, thank you $\endgroup$ Commented Nov 26 at 13:42
  • 1
    $\begingroup$ You can put $f_2 = 1/f_1$ to see that $f_1f_2$ can be everywhere differentiable while both $f_1$ and $f_2$ can be as badly behaved as you want. $\endgroup$ Commented Nov 26 at 13:50

1 Answer 1

Reset to default
5
$\begingroup$

The product rule isn't just an equation. It's a theorem, and that means it only works when its hypotheses, its preconditions, are satisfied.

Specifically, the theorem states that IF $f_1$ and $f_2$ both have derivatives at a point $x$ THEN the product rule formula works at that point. If only one or neither of $f_1$, $f_2$ have derivatives at $x$, then the product rule gives you nothing.

For example:

Let $f_1(x) = 2C+W(x)$ where $W$ is the nowhere-differentiable Weierstrass function and $C>0$ is a constant such that $|W(x)| \leq C$ everywhere, so this $f_1$ is everywhere continuous, nowhere differentiable, and nowhere zero. Let now $f_2(x) = \frac{1}{f_1(x)}$, also nowhere differentiable but defined and continuous everywhere.

Then $g(x)=1$ is differentiable everywhere.

$\endgroup$
1
  • $\begingroup$ (+1) Nice answer! $\endgroup$ Commented Nov 26 at 14:10

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.