Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Differentiability of composite functions

  1. Apr 28, 2012 #1
    Hi, I have a small question about this. Using the chain rule, I know that a composition of differentiable functions is differentiable. But is it also true that if a composition of functions is differentiable, then all the functions in the composition must be differentiable?

    For example, if [itex]f(g(x))[/itex] is differentiable, does that imply [itex]f(x)[/itex] and [itex]g(x)[/itex] are both differentiable?

    Thanks!
     
  2. jcsd
  3. Apr 28, 2012 #2

    [itex]\cos\sqrt{x}[/itex] is differentiable from the right at [itex]x=0[/itex] , but [itex]\sqrt{x}[/itex] isn't...

    DonAntonio
     
  4. Apr 28, 2012 #3

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Examples can be constructed where f(x) or g(x) can be arbitrarily bad (for example discontinuous) yet f(g(x)) is differentiable

    A classic example of this is when f(x)=1 and g(x) is any function you pick. f(g(x))=1 so this composition is differentiable, but g(x) clearly doesn't have to be.

    For the other way around consider f(x)="any function which is always negative" if x<1 (for example -x6), and x4 if x >=1. Now let g(x) = x2+3. Because g(x)>=1 always, f(g(x)) is always differentiable (because we are always using the x4 portion of f(x) thanks to how g was constructed), even though a large chunk of f(x) is not differentiable.
     
  5. Apr 30, 2012 #4
    Thanks for the help and examples!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook