5
$\begingroup$

How will be the product and composition of two functions, where one is differentiable and another is just continuous, behave?I mean to say, if the product or composition is differentiable, then what are the conditions on the functions to hold for their product to be differentiable, if they are?

Some trivial checks: Of course, the product/composition is not always differentiable since if we take the differentiable function to be I (or x), then the result is obviously not differentiable. So what I ask for is that when they do; why?

$\endgroup$
7
$\begingroup$

Let $f,g$ two functions and $h = f \circ g$, then

  • If $f$ and $g$ are differentiable, $h$ is also differentiable (see here)

  • If $h$ is differentiable, let $f(x)=|x|$ and $g(x)=x^2$, then $h(x)=x^2$ is differentiable on $\mathbb{R}$, but $f$ is not differentiable at $0$. Note that $g \circ f(x)=x^2$ is also differentiable. So, if $h$ is differentiable you can't say anything about $f$ and $g$ (another example is $f(x)=g(x)=\frac{1}{x}$, then $f$ and $g$ are not differentiable at $0$ but $h(x)=x$).

  • If $h$ is not differentiable, then either $f$ or $g$ (or both) is not differentiable (if not, this would contradicts the first point).

$\endgroup$
  • $\begingroup$ But f is not even defined at zero, right? $\endgroup$ – O_huck Jul 4 '14 at 16:13
  • $\begingroup$ $f\circ g$ is not defined at $0$... $\endgroup$ – Thomas Jul 4 '14 at 16:13
  • $\begingroup$ After edit: No $f\circ f$ is not $x^2$ $\endgroup$ – Thomas Jul 4 '14 at 16:14
  • $\begingroup$ The left hand limit is $-3x^2$ which gives 0 and right is $3x^2$ which also gives 0 at 0. $\endgroup$ – O_huck Jul 4 '14 at 16:16
  • 1
    $\begingroup$ @quarkine ?¿ $f'(x) = \text{sign}(x)$ which is not defined at $0$. $\endgroup$ – Surb Jul 4 '14 at 16:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.