Derivatives of trace of complicated function

Question

When $A$, $B$ and $C$ are positive semi-definite and hermitian symmetric matrices, what is the derivative and second derivatives of $f(x) = tr(x^2 A (xB + C)^{-1})$, for $x>0$ ?

Answer 1

• Consider $f(x) = x^2 A ( x B + C )^{-1}$. Then we have $$ f(x) (x B + C) = x^2 A $$ Differentiating on both sides yields $$ f(x) B + \frac{df}{dx}(x) (x B + C) = 2 x A $$ So we get $$ \frac{df}{dx}(x) = 2 x A (x B + C)^{-1} - x^2 A (x B + C)^{-1} B (x B + C)^{-1} $$
• The trace function is linear. So its derivative is itself.
• Next you combine these derivatives to get the composite function.

Answer 2

Define the matrix $$\eqalign{ M &= Ax^2(C+Bx)^{-1} \\ }$$ and calculate its differential and derivative $$\eqalign{ M(C+Bx) &= Ax^2 \\ M(B\,dx) + dM(C+Bx) &= 2Ax\,dx \\ dM &= (2Ax-MB)\,(C+Bx)^{-1}\,dx \\ P=\frac{dM}{dx} &= (2Ax-MB)\,(C+Bx)^{-1} \\ }$$ Write the function in terms of this new matrix.
Then calculate its derivatives. $$\eqalign{ f &= {\rm Tr}\big(M\big) \\ \frac{df}{dx} &= {\rm Tr}\big(P\big) \;=\; {\rm Tr}\Big((2Ax-MB)\,(C+Bx)^{-1}\Big) \\ \frac{d^2f}{dx^2} &= {\rm Tr}\Big( \left(2A+PB\right)(C+Bx)^{-1} - (2Ax-MB)\,(C+Bx)^{-1}B(C+Bx)^{-1} \Big) \\ }$$