Can you please help me with the derivative of $x'(A+tB)^{-1}x$ with respect to $t$?
($x$ is a vector, $A$ and $B$ are $n$ by $n$ matrices and $t$ is a scalar)
I'm not sure, but is it $x'Fx$ where $F=-(A+tB)^{-1}(B)(A+tB)^{-1}$?
[I applied $\frac{\delta C^{-1}}{\delta t} = -C^{-1}\frac{\delta C}{\delta t}C^{-1}$ Where here $C=(A+tB)$, the vectors $x$ remains as it is.]
Thanks in advance!
Looks like you're right. Working with differentials:
$\mathbb d(x^T(A + tB)^{-1}x)$ reduces to
$\mathbb d(x^T)((A + tB)^{-1}x) + x^T\mathbb d((A + tB)^{-1}x)$, and the left term cancels because $\mathbb d(x^T)$ is zero.
So you're left with
$x^T\mathbb d((A + tB)^{-1}x)$, which can be expanded to
$x^T\left(\mathbb d((A + tB)^{-1})x + ((A + tB)^{-1})\mathbb dx\right)$, and the rightmost term is zero because $\mathbb dx$ is zero
Now you're left with
$x^T\mathbb d((A + tB)^{-1})x$, where the middle term can be used to get $\mathbb dt$
= $x^T\left(-(A + tB)^{-1}(\mathbb dt)B(A + tB)^{-1}\right)x$
The derivative you want is everything in this expression excluding $\mathbb dt$, so it is
$x^T\left(-(A + tB)^{-1}B(A + tB)^{-1}\right)x$, which matches your answer.