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Suppose there are two positive definite quadratic forms $f(\mathbf x) = \mathbf x ^ T \mathbf C \mathbf x $ and $g(\mathbf x) = \mathbf x ^ T \mathbf G \mathbf x $. My goal is to get an expression for the derivative $$ \frac{\mathrm d}{\mathrm d \mathbf x} \left ( \frac {f(\mathbf x)}{g(\mathbf x)} \right ) .$$

Allegedly, $$ \frac{\mathrm d}{\mathrm d \mathbf x} \left ( \frac {f(\mathbf x)}{g(\mathbf x)} \right ) = \frac{\mathbf C \mathbf x f(\mathbf x) - \mathbf G \mathbf x f(\mathbf x)}{g(\mathbf x)^2} ,$$ although I don't know how to justify this result, or whether it is infact correct. Can somebody please show how to take this derivative. Please show a few intermediate steps, to facilitate understanding. Please do use $ \frac{\mathrm d}{\mathrm d \mathbf u} \left ( \mathbf u ^ T \mathbf A \mathbf u \right ) = 2 \mathbf u ^ T \mathbf A$ if appropriate.

EricVonB
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    You can rewrite $f(x)/g(x)$ as the quotient of two Rayleigh quotients $ \dfrac{\mathbf x ^ T \mathbf C \mathbf x / | \mathbf x |^2}{\mathbf x ^ T \mathbf G \mathbf x / | \mathbf x |^2}$. Indeed the derivative of a Rayleigh quotient is known: see (http://math.stackexchange.com/q/101528) where you can drop the "real parts" because you visibly work in $\mathbb{R}$. – Jean Marie Dec 30 '16 at 18:58
  • Your formula is erroneous: it should be : $\frac{\mathrm d}{\mathrm d \mathbf x} \left ( \frac {f(\mathbf x)}{g(\mathbf x)} \right ) = \frac{2 \mathbf C \mathbf x g(\mathbf x) - 2\mathbf G \mathbf x f(\mathbf x)}{g(\mathbf x)^2}$ – Jean Marie Dec 30 '16 at 19:12
  • I am not familiar with Rayleigh Quotients, so there is a lot there for me to digest. In the meantime, can you say whether the alleged derivative shown in the original posting is correct? – EricVonB Dec 30 '16 at 19:12
  • Have a look at: (http://fourier.eng.hmc.edu/e176/lectures/algebra/node9.html) – Jean Marie Dec 30 '16 at 19:14
  • That is a very useful link. Thank you. I think that is all I need. – EricVonB Dec 30 '16 at 19:18

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