Let $T$ be a bounded operator on a Banach space $X$. It's known that by the Dunford-Riesz functional calculus and the Cauchy formula we have that,
$$ f(T) = \frac{1}{2\pi i}\int_{\partial U} \frac{f(\lambda)} {\lambda - T} \, d\lambda, $$
where $f$ is an holomorphic function in some neighbourhood of the spectrum $\sigma(T)$ of $T$ and $U$ is an open set such that $\sigma(T) \subset U$ and $\partial U$ consists of a finite number of rectificable Jordan curves oriented in positive sense.
On the other hand from complex analysis we have Cauchy's differentiation formula that states that if $f$ is an holomorphic complex function,
$$ f^{(n)}(a) = \frac{n!}{2\pi i}\int_{\Gamma} \frac{f(z)} {(z - a)^{n+1}} \, dz, $$ where $\Gamma$ is a rectificable Jordan curve.
My question is the following, can we say that this inequality holds also for bounded operators? That is, is it true that,
$$ f^{(n)}(T) = \frac{n!}{2\pi i}\int_{\partial U} \frac{f(\lambda)} {(\lambda - T)^{n+1}} \, d\lambda. $$
The way I see it is that it has to hold considering the function $f^{(n)}$ in the Dunford-Riesz functional calculus and then do some integration by parts. The problem is that as we are working with a Bochner integral, i don't know how to properly continue.
Any reference is welcome as well!
We will follow the construction shown in the book [1], in particular we follow section 1.9 The Riemann-Stieltjes Integral and we will use 2 results:
First result
Formula (1.20)
It follows that if $g$ is Riemann-Stieltjes integrable with respect to $F$, then so is $F$ with respect to $g$ (and vice versa, by symmetry) and the following formula of integration by parts holds:
$$ \int_a^b g(t) d F(t)=g(b) F(b)-g(a) F(a)-\int_a^b F(t) d g(t). $$
Second result
Proposition 1.9.11
Let $g:[a, b] \rightarrow \mathbb{C}$ and $F:[a, b] \rightarrow X$. If $F$ is an antiderivative of a Bochner integrable function $f$ and if $g$ is continuous, then $\int_a^b g(s) d F(s)$ exists and equals the Bochner integral $\int_a^b g(s) f(s) d s$. If $F$ is continuous and $g$ is absolutely continuous, then $\int_a^b F(s) d g(s)$ equals the Bochner integral $\int_a^b F(s) g^{\prime}(s) d s$.
Answer to the question
We will restrict to the case where $\partial U$ is a circunference that is $\partial U = \{\lambda \in \mathbb{C} \mid \lvert \lambda \rvert = r\}$ for some $r>1$. Therefore by the Dunford-Riesz Calculus we have that,
$$ f'(T) = \frac{1}{2\pi i} \int_{-\pi}^\pi \frac{ri f'(r e^{it})}{r e^{it} - T} \, dt. $$
Calling $F(t) = \frac{1}{r e^{it} - T}$ and $g(t) = f(r e^{it})$ we have that,
$$ f'(T) = \frac{1}{2\pi i} \int_{-\pi}^\pi F(t) dg(t), $$
where, if we apply integration both results mentioned above we have that,
$$ \begin{aligned} f'(T) & = \frac{1}{2\pi i} \int_{-\pi}^\pi F(t) dg(t) = -\frac{1}{2\pi i} \int_{-\pi}^\pi g(t) dF(t) \\ & = \frac{1}{2\pi i} \int_{-\pi}^\pi \frac{r i f(r e^{it})}{(r e^{it} - T)^2}\, dt = \frac{1}{2\pi i} \int_{\partial U} \frac{f(\lambda)}{(\lambda - T)^2}\, d\lambda. \end{aligned} $$
In order to obtain the general formula we just have to keep doing this trick.
References
[1] Arendt, W., Batty, C., Hieber, M., & Neubrander, F. (2011). Vector-valued Laplace transforms and Cauchy problems. (Vol. 96) Birkhäuser/Springer Basel AG, Basel.