Differentiability of composite functions

  • Thread starter raphile
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  • #1
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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!
 

Answers and Replies

  • #2
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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!

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

DonAntonio
 
  • #3
Office_Shredder
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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.
 
  • #4
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Thanks for the help and examples!
 

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