I add my thoughts to the already existing fine answers.
1. The answer is "no" as follows from Propositions 79 and 92 in Tao's lecture notes on differentiation theorems. Indeed, let $f:[a,b]\to\mathbb{R}$ be an increasing differentiable function. By the quoted propositions, $f'$ is absolutely integrable, and $$f(x)=f(a)+\int_a^x f'(t)\,dt,\qquad x\in[a,b].$$ Therefore, by a well-known criterion (cf. #6 of Exercise 87 in the notes), $f$ is absolutely continuous.
2. Here is a more conceptual explanation based on Theorems 6.10, 6.11, 7.14, 7.21 in Rudin: Real and complex analysis (1987). Consider the Lebesgue-Stieltjes measure $df(x)$ associated with $f$, and consider its Lebesgue decomposition as in Theorem 6.10. By Theorem 7.14, the absolutely continuous part is $f'(x)dx$, which by Theorem 7.21 is the whole $df(x)$ (i.e. the singular part is zero). So $df(x)$ is absolutely continuous with respect to $dx$, which by Theorem 6.11 means that $f$ is absolutely continuous. See also Theorem 7.18 for a slight variation.