Is the fundamental theorem of calculus independent of ZF?
I believe that one of the standard proofs works.
Let $F(x) := \intop_a^x f^\prime (t) dt$. Then $F$ is differentiable and its derivative is $f^\prime$ due to a standard estimate that has nothing to do with AC.
$(F-f)^\prime = 0$, hence it is constant. This boils down to the one-dimensional case: just consider $g := \Vert F-f-F(a)+f(a) \Vert$. It is a real-valued function with zero derivative, and $g(a)=0$, so we can use the usual "one-dimensional" mean value theorem.
Claim: Let $g:[a,b]\to B$ be differentiable, with $g'(t)=0$ for all $t\in[a,b]$. Then $g(t)$ is a constant.
Proof: Fix $\epsilon>0$. For each $t\in[a,b]$, we can find $\delta_t>0$ such that $0<|h|<\delta_t\Rightarrow\|g(t+h)-g(t)\|<\epsilon|h|$. The open intervals $(t-\delta_t,t+\delta_t)$ cover $[a,b]$, so there is a finite subcover $\{(t_i-\delta_{t_i},t_i+\delta_{t_i}):1\leq i\leq N\}$. We may choose our labeling so that $t_1<t_2<\ldots<t_N$. Now, we should be able to find points $x_0,\ldots,x_N$ with $x_0=a<t_1<x_1<t_2<\ldots<t_N<x_N=b$, satisfying $|x_i-t_i|<\delta_i$ and $|x_{i-1}-t_i|<\delta_i$ for $1\leq i\leq N$. Now, \begin{eqnarray*} \|g(b)-g(a)\|&=&\left\|\sum_{i=1}^N(g(x_i)-g(t_{i}))-(g(t_i)-g(x_{i-1}))\right\|\\ &<&\sum_{i=1}^N \epsilon (x_i-t_i)+\epsilon(t_i-x_{i-1})\\ &=&\epsilon(b-a) \end{eqnarray*} Since $\epsilon$ is arbitrary, we have $g(b)=g(a)$. This argument works on any subinterval, so $g(t)$ is constant.
We can apply the above with $g(x):=\int_a^xf'(t)\,dt-(f(x)-f(a))$. I believe it can be checked directly that $g'(x)=0$ for all $x$, and that $g(a)=0$, so that $g(x)$ must be identically 0.