Let $\varphi:\Bbb R\to\Bbb R$ be continuously differentiable and $g:[a,b]\to\Bbb R$ be continuous. Are we able to show that $$\varphi(g(b))-\varphi(g(a))=\varphi'(g(x_0))(g(b)-g(a))\tag1$$ for some $x_0\in[a,b]$? This claim is made in the proof of Theorem 3.70 of PDE and Martingale Methods in Option Pricing and it is stated that it would follow from the mean value theorem. However, it's obviously not an immediate application of the mean value theorem.
So, how do we obtain it?
I first thought we might need to use that $I:=g([a,b])$ is again a closed interval and then apply the mean value theorem to the restriction of $\varphi$ to $I$. However, this only yields the existence of $x_0\in I$ with $\varphi'(x_0)=\frac{\varphi(\max I)-\varphi(\min I)}{\max I-\min I}$, which is obviously not the same as $(1)$ (unless $g$ is nondecreasing, which is not assumed in the reference).
I'd also be interested in a generalization of the same claim to the case when $\varphi:E\to F$ and $g:[a,b]\to E$ for some $\Bbb R$-Banach spaces $E$ and $F$. In this case, there should be some kind of mean-value inequality yielding the existence of $x_0\in[a,b]$ with $$\left\|\varphi(g(b))-\varphi(g(a))\right\|_F\le\left\|{\rm D}\varphi(g(x_0))(g(b)-g(a))\right\|_F\tag2$$ or at least $$\left\|\varphi(g(b))-\varphi(g(a))\right\|_F\le\left\|{\rm D}\varphi(g(x_0))\right\|_{\mathfrak L(E,\:F)}\left\|g(b)-g(a)\right\|_E\tag3.$$
If $g(a)=g(b)$, $LHS=RHS=0$ for any choice of $x_{0}\in[a,b]$. Suppose that $g(a)\neq g(b)$. If $g(a)<g(b)$, we apply Mean-Value Theorem on $\varphi\mid_{[g(a),g(b)]}$ (the restriction of $\varphi$ on $[g(a),g(b)]$), there exists $\xi\in(g(a),g(b))$ such that $\varphi(g(b))-\varphi(g(a))=\varphi'(\xi)\left[g(b)-g(a)\right]$. Since $g$ is continuous, by Intermediate Value Theorem, we can choose $x_{0}\in(a,b)$ such that $g(x_{0})=\xi$. That is, $\varphi(g(b))-\varphi(g(a))=\varphi'(g(x_{0}))\left[g(b)-g(a)\right]$.
The case that $g(a)>g(b)$ can be treated similarly.