Consider the following statement:
Consider two functions $f:[a,b]\to \mathbb{R}$ and $\phi:[c,d]\to [a,b]$ with the following properties:
1)$f$ is continuous
2)$\phi$ is bijective, $\phi$ and $\phi^{-1}$ are differentiable and their derivatives are continuous.
Then $$\int_a^b f(x)dx=\int_{\phi^{-1}(a)}^{\phi^{-1}(b)}f(\phi(t))\phi'(t)dt.$$
The proof given in my textbook goes as follows:
Let $F:[a,b] \to \mathbb{R}$ be an antiderivative of $f$ (it exists since $f$ is continuous). Then $\int_a^b f(x)dx=F(b)-F(a)(1)$.
By the chain rule, we have that $(F\circ \phi)'(t)=F'(\phi(t))\phi'(t)=f(\phi(t))\phi'(t)$, which shows that $F\circ \phi$ is an antiderivative of $(f\circ \phi)\cdot \phi'$.
As a result, $\int_{\phi^{-1}(a)}^{\phi^{-1}(b)}f(\phi(t))\phi'(t)dt=(F\circ \phi)(t) |_{\phi^{-1}(a)}^{\phi^{-1}(b)}=F(b)-F(a)(2)$.
From $(1)$ and $(2)$ we reach the desired conclusion.
My question is: why is it necessary that $\phi$ and $\phi^{-1}$ have continuous derivatives? Wouldn't it be enough if they were just Riemann integrable?