# Derivative of ratio of functions, $f,g:\mathbb{R}^N\to\mathbb{R}$ with respect to a vector?

This is fairly simple, but my matrix calculus is not that strong. Given two functions, $f:\mathbb{R}^N\to\mathbb{R}$, $g:\mathbb{R}^N\to\mathbb{R}$, and $x\in\mathbb{R}^N$, how do I compute the following derivative $$\frac{\partial }{\partial x}\left(\frac{f(x)}{g(x)}\right)$$

In the scalar case ($N=1$) the derivative of the ratio is just obtained by the basic product rule $$\frac{\partial }{\partial x}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)}{g(x)} - \frac{f(x)g'(x)}{g^2(x)}$$ But if I try to generalize this to the multivariate case, the derivative terms are no longer scalars, they become vectors of dimension $N\times 1$. Does the above idea of applying the product rule generalize to the multivariate case? I realize there needs to be some transpose operators somewhere in there but I'm not sure where.

Also, how would I write out the second derivative in the multivariate case? $$\frac{\partial^2 }{\partial x^2}\left(\frac{f(x)}{g(x)}\right)$$

(any good references on matrix calculus to help me with this stuff is appreciated!)

Thank you!

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The ratio rule holds as you have written. It is just that $f'$ and $g'$ have to be replaced with the corresponding gradients. – Fabian Jan 1 '13 at 22:58

You just need a little bit of the chain rule and the product rule.

The vector derivative still obeys the product rule:

$$\nabla \frac{f}{g} = \frac{1}{g} \nabla f + f \nabla \frac{1}{g}$$

The key is doing the derivative of $1/g$. We can set $\tau = 1/g$ and see that $d\tau/dg = -1/g^2$. The key then is to use the chain rule to get

$$\nabla \tau = \frac{d\tau}{dg} \nabla g$$

As has been said, this essentially reproduces the single-variable result. The Laplacian can be attacked in a similar way. The keys to this approach are just to use the product rule, along with the chain rule to turn division into multiplication.

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Thanks for the help. Just to make sure I am computing the second derivative correctly, does someone mind checking the work below? So, we know that $$\nabla\frac{f}{g} = \frac{1}{g}\nabla f + f\nabla \frac{1}{g} = \frac{1}{g}\nabla f - \frac{f}{g^2}\nabla g$$. This is an $N$-dimensional vector. Now the second derivative is given by, $$\nabla^2 \frac{f}{g} = \frac{1}{g}\nabla^2 f - \frac{1}{g^2}\nabla g\nabla f - \frac{f}{g^2}\nabla^2 g - \frac{1}{g^2}\nabla f\nabla g + \frac{2f}{g^3}\nabla g\nabla g$$ But what does $\nabla f\nabla g$ mean, both are col. vectors? Need a transpose? – Erik M Jan 4 '13 at 3:21
Yes, terms that are $\nabla g \nabla f$ should be written as $\nabla g \cdot \nabla f$, which means that, interpreted as matrix operations, it would be $(\nabla g)^\dagger \nabla f$. – Muphrid Jan 4 '13 at 3:34
Thanks! But should the tranpose operator be on the $\nabla f$ term instead of the $\nabla g$ term, since we know $\nabla^2 \frac{f}{g}$ should be a matrix? (assuming both are column vectors) – Erik M Jan 4 '13 at 3:40
It kind of depends what you mean by $\nabla^2$. In vector calculus, $\nabla^2$ is a scalar differential operator--in that it takes scalar fields to scalar fields. What you want could be written in index notation as $H_{ij} = (e_i \cdot \nabla)(e_j \cdot \nabla) f/g$, where the indices $i,j$ denote the corresponding entry in the matrix $H$. $e_i, e_j$ are basis vectors. – Muphrid Jan 4 '13 at 4:31