Differentiability of composite functions

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Discussion Overview

The discussion revolves around the differentiability of composite functions, specifically whether the differentiability of a composition implies the differentiability of its constituent functions. Participants explore examples and counterexamples related to this question.

Discussion Character

  • Debate/contested

Main Points Raised

  • One participant questions if the differentiability of a composition, such as f(g(x)), necessitates that both f(x) and g(x) are differentiable.
  • Another participant provides an example where cos(√x) is differentiable from the right at x=0, while √x is not differentiable at that point.
  • It is noted that examples can be constructed where either f(x) or g(x) can be discontinuous or otherwise non-differentiable, yet f(g(x)) remains differentiable. For instance, if f(x)=1 and g(x) is any function, then f(g(x))=1 is differentiable regardless of g(x).
  • Another example is presented where f(x) is defined piecewise, being non-differentiable for x<1 and differentiable for x≥1, while g(x) is always greater than or equal to 1, ensuring that f(g(x)) is differentiable despite f(x) having non-differentiable segments.

Areas of Agreement / Disagreement

Participants express differing views on whether the differentiability of a composition implies the differentiability of its components. Multiple competing examples and arguments are presented, indicating that the discussion remains unresolved.

Contextual Notes

Participants highlight the need for careful consideration of definitions and the specific conditions under which differentiability is evaluated, particularly in the context of piecewise functions and discontinuities.

raphile
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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 f(g(x)) is differentiable, does that imply f(x) and g(x) are both differentiable?

Thanks!
 
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raphile said:
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 f(g(x)) is differentiable, does that imply f(x) and g(x) are both differentiable?

Thanks!


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

DonAntonio
 
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.
 
Thanks for the help and examples!
 

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