Differentiability of composite functions

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The discussion addresses the differentiability of composite functions, specifically whether the differentiability of a composition implies that all individual functions within it are also differentiable. It is established that while the chain rule confirms that the composition of differentiable functions is differentiable, the reverse is not necessarily true. Examples illustrate that a composite function can be differentiable even if one or both of its constituent functions are not. For instance, a constant function can yield a differentiable composition regardless of the behavior of the other function. Overall, the key takeaway is that differentiability of a composite function does not guarantee the differentiability of its components.
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!
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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