# Vector Matrix Differentiation (to maximize function)

how would I calculate the derivative of the following. I want to know the derivative so that I can maximise it.

$$\frac{x^TAx}{x^TBx}$$

Both the matricies A and B are symmetric. I know the derivative of $\frac{d}{dx}x^TAx = 2Ax$. Haven't been very successful applying the quotient rule to the above though. Appreciate the help. Thanks!

EDIT: In response to "What goes wrong when applying the chain rule". We know that: $$\frac{d}{dx}\frac{u}{v} = \frac{vu' - uv'}{v^2}$$ Which would give me: $$\frac{2x^TBxAx - 2x^TAxBx}{x^TBx^2} \, or \, \frac{2Axx^TBx - 2Bxx^TAx}{(x^TBx)^2}$$

In the first case the dimensions don't agree. In the second they do, but I don't want to assume that it's correct just because the dimensions agree. If it is correct then please do let me know!

-
What goes wrong when you try to apply the chain rule? –  Owen Biesel Oct 13 '12 at 4:47
See edit. Thanks –  foges Oct 13 '12 at 4:57

Both expressions are correct, up to some missing parentheses, in the following sense: For each $i$, we have $\frac{\partial}{\partial x_i}(x^TAx) = 2(Ax)_i$ and $\frac{\partial}{\partial x_i}(x^TBx) = 2(Bx)_i$, so the quotient rule tells us that

$$\frac{\partial}{\partial x_i}\frac{x^TAx}{x^TBx} = \frac{2(Ax)_i(x^TBx) - 2(Bx)_i(x^TAx)}{(x^TBx)^2}.$$

This may be written, if you interpret $d/dx$ as gradient, as

$$\frac{d}{dx}\frac{x^TAx}{x^TBx} = \frac{2Ax(x^TBx) - 2Bx(x^TAx)}{(x^TBx)^2} = \frac{2Axx^TBx - 2Bxx^TAx}{(x^TBx)^2},$$ so your second expression is correct.

Since $(x^TAx)$ and $(x^TBx)$ are just numbers, if left inside their parentheses they can multiply a vector or matrix from either side. Hence the first expression is also correct if you are sure to leave in the parentheses:

$$\frac{d}{dx}\frac{x^TAx}{x^TBx} = \frac{2(x^TBx)Ax - 2(x^TAx)Bx}{(x^TBx)^2}.$$

-
Ok great, thanks. That makes sense :) –  foges Oct 13 '12 at 5:27