Let $f: R^n \to R$ be given by $f(x) = \frac{||x||^4} {1 + ||x||^2}$ . Use the chain rule to show that $f$ is differentiable at each $x \in R^n$ and compute $Df(x)$.
This vector valued stuff just came out of nowhere. It's not in our book and we went through it super fast... so I, and many others, are completely lost. From what I can guess, each $x$ is a vector of length "n", and $||x||$ is supposed to be the norm... Since the question doesn't specify, I'm assuming it's the standard euclidean norm. Also, $D$ must be an n*n matrix of partial derivatives? Assuming all of that is correct, I've been sitting trying to figure out how to make use of the chain rule, but I've got nothing :/
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$\begingroup$Let's recap some things: For general $f=(f_1,\ldots,f_m):U\subseteq\mathbb{R}^n\to\mathbb{R}^m$ (where $U$, the domain of $f$, is open in $\mathbb{R}^n$) and $x\in U$, $Df(x)$ is the Jacobian (or rather the linear transformation associated to it), given by $$Df(x)=\left[\frac{\partial f_i}{\partial x_j}(x)\right]_{\substack{i=1,\ldots,m\\j=1,\ldots,n}}$$ (wherever this makes sense). From this definition, the following are obvious:
When $m=1$, $Df(x)$ is simply a row, and in fact it is equal to $\nabla f(x)$, the gradient of $f$ at $x$.
When $n=1$, $Df(x)$ is a column, where each entry is simply the derivative of $f_j$ at $x$.
When $m=n=1$, $Df(x)$ is a number, equal to $f'(x)$.
Now recall the Chain Rule:
For $f:U\subseteq \mathbb{R}^n\to V\subseteq\mathbb{R}^m$ and $g:V\subseteq\mathbb{R}^m\to\mathbb{R}^p$, $$D(g\circ f)(x)=Dg(f(x))Df(x),$$ where the RHS is simply product of matrices.
Now to your problem: to use the chain rule, you have to see $f$ as a composition of two simpler functions, which are differentiable everywhere. If we put $h:\mathbb{R}^n\to(0,\infty)$ as $h(x)=\Vert x\Vert^2$, and $k:(0,\infty)\to\mathbb{R}$ by $k(t)=\frac{t^2}{1+t}$, we have $f=k\circ h$, and it is easier to calculate $Dh$ and $Dk=k'$.
I'll leave the details to you, but here's what you should get: For $x=(x_1,\ldots,x_n)$, use the definition of $Dh(x)$ to find $Dh(x)=2[x_1\ x_2\cdots x_n]$.
For $k$, one-variable calculus gives $k'(t)=\frac{t^2+2t}{(1+t)^2}$.
By the chain rule, $$Df(x_1,\ldots,x_n)=2\frac{(\Vert x\Vert^4+2\Vert x\Vert^2)}{(1+\Vert x\Vert^2)^2}[x_1\cdots x_n]$$
$\endgroup$ 2 $\begingroup$I assume that you are working with the euclidean norm. You are right that these are length n (1 by n or n by 1 depending on convention) vectors. So we have by definition of the norm
$f(x)=\frac{||x||^4}{1+||x||^2}=\frac{\sqrt{(x_{1}^{2}+x_{2}^{2}+....+x_{n}^{2})}^4}{1+\sqrt{(x_{1}^{2}+x_{2}^{2}+....+x_{n}^{2})}^2}= \frac{(x_{1}^{2}+x_{2}^{2}+....+x_{n}^{2})^2}{1+(x_{1}^{2}+x_{2}^{2}+....+x_{n}^{2})}$
Just reducing exponents. Can you take it from here using the chain rule?
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