TITLE: $\frac{d}{dt} (A+t B)^p\,\text{ for } p\geq 1$
QUESTION [13 upvotes]: Given two positive self-adjoint operators $A,B$ on a Hilbert space. Let $p\geq 1$.
I would like to calculate $$\frac{d}{dt}|_{t=0} (A+tB)^p,$$
where the power is defined through the spectral theorem. The difficulty is that $A$ and $B$ do not commute in general. Is there a nice formula for that? 
I thought about using some integral representation of $x^p$, which hopefully would make the differentiation easier, but I was not successful. 
Update:
Thanks for your answers. In my case, I actually only need it for $p\in [1,2)$. Using $y^p = \frac{\sqrt 3}{2\pi}\int_{0}^\infty dx \,\frac{y^2 x^{p-2}}{x+y}$, I thought about writing $C:=A+tB$ as $$C^p = \frac{\sqrt 3}{2\pi}\int_0^\infty C^2 x^{p-2} (x+C)^{-1} dx.$$
That might also work. I am still happy about other suggestions.

REPLY [4 votes]: I give you the answer in the finite dimensional case, and let you adapt it to the Hilbert space case. Let $f$ be an analytic function; in your case $f(t)=t^p$.
First of all, choose an orthonormal basis in which $A=:D$ is diagonal. Then form the matrix
 $f^{[1]}(D)\in{\bf M}_n({\mathbb C})$ by
$$f^{[1]}(D)_{jk}=\left\{\begin{array}{lr}
f'(d_j), & k=j, \\ \\ \frac{f(d_j)-f(d_k)}{d_j-d_k}, & k\ne j,\end{array}\right.$$
where we identify
$$\frac{f(b)-f(a)}{b-a}:=f'(a),$$
if $b=a$. Then the differential of $H\mapsto f(H)$, taken at $D$, is
$${\rm D}f(D)B=f^{[1]}(D)\circ B,
$$
where $A\circ B$ denotes the Hadamard product.
Edit. As noticed by Suvrit, this is Daleckii-Krein Formula.