University Calculus: Early Transcendentals (3rd Edition)

We revisit what is claimed in the Chain Rule: - If $f(u)$ is differentiable at $u=g(x)$. - If $g(x)$ is differentiable at $x$. $\Rightarrow$ $(fog)(x)$ is differentiable at $x$. But it does not claim the way around. In other words, if $f(u)$ is differentiable at $u=g(x)$ and $g(x)$ is differentiable at $x$, then $(fog)(x)$ is definitely differentiable at $x$; but if $(fog)(x)$ is differentiable at $x$, then it is not necessary that $f(u)$ must be differentiable at $u=g(x)$ or $g(x)$ must be differentiable at $x$. The Chain Rule is not a rule for the conditions of differentiability. Applying to this exercise, we see an example that while $(fog)(x)$ and $(gof)(x)$ are both differentiable at $x=0$, it does not have to be true that $g(x)$ must be differentiable at $x=0$ also. In conclusion, this case does not contradict the Chain Rule.