Exercise 2.5.10(b) and the converse of the chain rule

[mikeshulman]

Exercise 2.5.10(b) asks to find a value of x where C'(x) does not exist, where C(x) = p(q(x)). The solution says

Consider C′(1). By the chain rule, we'd expect that C′(1)=p′(q(1))⋅q′(1), but we know that q′(1) does not exist since q has a corner point at x=1. This means that C′(1) does not exist either.

However, this is actually invoking a converse of the chain rule (or, I suppose, more precisely an inverse, but that's equivalent), which is significantly subtler. According to this math.se question and its answer, in certain circumstances one can conclude the differentiability of g(x) at a, or that of f(u) at g(a), from the differentiability of f(g(x)) at a; but only when the other function is also differentiable at its corresponding point, with a nonzero derivative, and certain continuity assumptions hold. These additional assumptions do hold in the situation of 2.5.10(b), but we certainly don't want to discuss these complicated statements in a calculus 1 course, and it doesn't seem to me like a good idea to encourage students to conflate a theorem with its converse.

One possibility would be to modify the exercise to ask "find a value of x for which the chain rule does not allow you to conclude that C'(x) exists" or "...does not allow you to find a value for C'(x)".