I believe the problem of trying to find the Jacobian of the following function highlights a lack of understanding of some concept on my part. I was hoping someone could either provide specific advice about solving this problem, or computing Jacobians in general.
Consider the mapping $h : \mathbb{R}^n \rightarrow \mathbb{R}^n$ where the domain is length-$n$ column vectors and the range length-$n$ row vectors (or a transposed vector, if you like). The function is
$$h(x) = \frac{\eta v' + (M x)'}{(\eta + u'x)^2},$$
where the constants $v$ and $u$ are (column) vectors, $\eta$ is a scalar, and $M$ is a square matrix.
So far as I know, the quotient rule for vectors is
$$\nabla\left(\frac{f}{g}\right) = \frac{g\nabla f – f \nabla g}{g^2}$$
and
\begin{align*}
\nabla f &= M'\\
\nabla g &= 2(\eta + u'x) u'
\end{align*}
Putting it all together, I get
$$\nabla h = \frac{(\eta + u'x)^2 M' – [\eta v' + (M x)'] 2(\eta + u'x) u'}{(\eta + u'x)^4}.$$
This expression is clearly not right, and to see why evaluate the Jacobian at $x = \mathbf{0}$:
$$\nabla h(0) = \frac{\eta^2 M' – 2\eta^2 v' u'}{\eta^4}$$
The resulting expression should be a $n \times n$ matrix, but in the second term we have two (row) vectors multiplied by one another. It seems likely there should be some sort of outer product here, but I'm not sure where my math is going wrong.
Any help you can provide is greatly appreciated.
Best Answer
My suggestion for minimizing notational confusion is to focus on directional derivative, in the direction of some fixed vector $w$. This derivative $D_w h$ is always a function of the same nature as $h$: here, it will also eat column vectors and spit out row vectors. For the numerator the computation is easy since it's linear: $$D_w(\eta v' + (M x)') = (Mw)' \tag1$$ while $(\eta + u'x)^{-2}$ has, by the chain rule, $$D_w((\eta + u'x)^{-2}) = -2 (\eta + u'x)^{-3} D_w(\eta + u'x) =-2 (\eta + u'x)^{-3} (u'w) \tag2$$ Combine (1) and (2) by the product rule: $$D_w h = -2 (\eta + u'x)^{-3} (u'w) (\eta v' + (M x)') +(\eta + u'x)^{-2}(Mw)' \tag3 $$ As promised, the right hand side of (3) is a row vector. It depends linearly on $w$ and thus defines a $(1,1)$ tensor field.