Computing the derivative of an inverse matrix

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To compute the derivative of the inverse matrix (A+tB) at t=0, the expression is derived as -(A+tB)^{-1} * d/dt(A+tB) * (A+tB)^{-1}. Evaluating at t=0 requires simplifying the right-hand side, which involves treating d/dt as a linear operator. The derivative of A is zero since it is constant, while the derivative of tB is simply B. An alternative method involves rewriting A+tB as A(I+t*A^{-1}B) and using power expansion, which leads to a similar result.
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Homework Statement


If A, B are elements of Mat(n, R) and A is invertible, compute

\frac{d}{dt}_{t=0}(A+tB)^{-1}


The Attempt at a Solution



The derivative will be of the form

\frac{d}{dt}(A+tB)^{-1}=-(A+tB)^{-1}\frac{d}{dt}((A+tB))(A+tB)^{-1}

but I need to evaluate this at t=0, so how do I simplify the expression on the right hand side? Since d/dt is a linear operator do I just attack each term individually, that is, take a derivative of A with respect to t plus a derivative of tB with respect to t?

This really is hard notation for me to follow.
 
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Just take the derivative in the middle and set t=0 on the outside factors. Another way to do it would be by writing A+tB as A(I+t*A^(-1)*B), taking the inverse and power expanding (I+t*A^(-1)*B)^(-1), but I seem to get the same thing.
 
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Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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