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Determine if the following statement is true or false. Provide proof if true, or a counterexample if false.
Statement:
If $(f\circ g)(x)$ is differentiable, then $f(x)$ and $g(x)$ must be differentiable.
I think that this statement is false and my counterexample is as follows: let $f(x) = |x|$, $g(x) = x^2$, however, I wanted to find a different example that doesn't make use of $x$, $x^2$, or any constant functions, but I can't seem to find one.
Could anybody give me a hand, please?
Take any bijection $f:\mathbb R\to\mathbb R$, and consider its inverse. Then $f\circ f^{-1}$ is the identity function which is obviously differentiable, but $f$ need not be.
$$ f(x) =g(x)= \begin{cases} 1, & \text{if $x$ is rational} \\ 0, & \text{if $x$ is irrational} \end{cases}$$
I wanted to mention this even though $f\circ g$ is constant because it seems cool to me that neither $f$ nor $g$ are continuous anywhere.
Let $\text{sign}(x) = 1$ if $x >= 0$, and $-1$ if $x<0$
Let $f(x) = -\text{sign}(x)$
Let $g(x) = 1$ if $x<0$ else $g(x)=x$
Then $(f \circ g)(x) = 1$
