Let $f: \mathbb{H} \rightarrow \mathbb{R}$ be a differentiable mapping, where $\mathbb{H}$ is a banach space. Consider the segment $tx$. This way, we have a function $\tilde{f}:[0,1] \rightarrow \mathbb{R}$ given by $\tilde{f}(t)=f(tx)$.
By the fundamental theorem of calculus,
$$f(x)=f(0)+\int_0^1 \tilde{f}'(t)dt=f(0)+\int_0^1 D_{tx}f(x)dt$$
We shall demonstrate the following lemmas:
Lemma: Let
$$A: \mathbb{H} \rightarrow L(\mathbb{H},\mathbb{R})$$
$$x \mapsto A_x$$
and
$$g: \mathbb{H} \rightarrow \mathbb{H}$$
be differentiable functions. Then,
$$D_x(A(g))=A_x(D_xg(~ \cdot ~))+ (D_xA(~\cdot ~)) g(x)$$
Proof:
$$A_{x+h}(g(x+h))=(A_x+D_xA(h)+\epsilon(h))(g(x)+D_xg(h)+\xi(h))$$
$$=A_xg(x)+A_x(D_xg(h))+(D_xA(h))(g(x))+ \zeta(h)$$
where $\frac{\zeta(h)}{||h||} \rightarrow 0$
$\blacksquare$
Corollary: [Integration by Parts] Given a function $\xi :\mathbb{R} \rightarrow \mathbb{H}$ and two functions $A$, $g$ as in the previous lemma, we have:
$$A_{\xi(1)}(g(\xi(1)))-A_{\xi(0)}(g(\xi(0)))=\int_0^1 A_{\xi(t)}(D_{\xi(t)}g( \xi'(t))) dt + \int_0^1 (D_{\xi(t)}A(\xi '(t)))(g(\xi(t)))dt$$
Proof
Fundamental Theorem of Calculus.
$\blacksquare$
Corollary: Given differentiable functions:
$$A: \mathbb{H} \rightarrow L(\mathbb{H},\mathbb{R})$$
$$x \mapsto A_x$$
and
$$h: \mathbb{R} \rightarrow \mathbb{H}$$
,we have:
$$A_{h(1)}(h(0))-A_{h(0)}(h(1))=\int_0^1 A_{h(t)}(-h'(t)) dt + \int_0^1 (D_{h(t)}A(h '(t)))(h(1)+h(0)-h(t))dt$$
Proof: Use $\xi=h$ and $g(z)=h(1)+h(0)-z$ in the previous corollary.
$\blacksquare$
Applying the above corollary to $h(t)=(1-t)x$ and $A=Df$, we obtain:
$$D_0f(x)-D_xf(0)=\int_0^1D_{(1-t)x}f(x)dt+\int_0^1((D_{(1-t)x}Df)(-x))(x-(1-t)x)dt$$
We then have:
$$\int_0^1D_{(1-t)x}f(x)dt =D_0f(x) -\int_0^1((D_{(1-t)x}Df)(-x))(x-(1-t)x)dt$$
Changing variables, we get:
$$\int_0^1D_{tx}f(x)dt =D_0f(x) -\int_0^1((D_{(tx)}Df)(-x))(x-tx)dt$$
$$\implies \int_0^1D_{tx}f(x)dt =D_0f(x) - \int_0^1((D_{(tx)}Df)(-x))((1-t)x)dt$$
Therefore,
$$f(x)=f(0)+D_0f(x)- \int_0^1((D_{(tx)}Df)(-x))((1-t)x)dt$$
Note that, in the above equation, we have the same hypotheses of the first lemma. This way, we can keep applying integration by parts repeatedly, obtaining:
$$f(x)=f(0)+D_0f(x)+\frac{1}{2}(D_0Df(x))(x)+...+R$$
where $R$ is an integral remainder
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