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The problem goes like this:

If $f:\mathbb{R}^n\to\mathbb{R}, f(x)=\arctan||x||^4$, prove that $Df(x)(x)=\displaystyle\frac{4||x||^4}{1+||x||^8}$

Now, I've calculated each of the partial differentials (if that's the right word) and applied that $1\times n$ matrix to a vector $(x_1, ... ,x_n)$ and I get this:

$4(\displaystyle\frac{x_1^4}{1+x_1^8}+...+\frac{x_n^4}{1+x_n^8})$

Now, the similarity between those terms and the final solution is obvious but I just can't seem to get the sum to become the above.

Am I going about this the wrong way or am I just missing something?

($||\cdot||$ is the Euclidian norm)

Luka Horvat
  • 2,618

1 Answers1

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$\mathrm{Arctan}$ is differentiable on $\mathbb{R}$ with :

$$ \big( \mathrm{Arctan}' \big)(x) = \frac{1}{1+x^{2}} $$

And $\varphi \, : \, x \in \mathbb{R}^{n} \, \longmapsto \, \Vert x \Vert^{4}$ is differentiable on $\mathbb{R}^{n}$ with :

$$ \mathrm{D}_{x}\varphi \cdot h = 4 \Vert x \Vert^{2} \left\langle x,h \right\rangle $$

where $\left\langle \cdot,\cdot \right\rangle$ denotes the usual inner product on $\mathbb{R}^{n}$. Since $f = \mathrm{Arctan} \circ \varphi$, the chain rule gives you :

$$ \begin{align*} \mathrm{D}_{x}f \cdot h & = {} \frac{1}{1+\big( \Vert x \Vert^{2} \big)^{2}} \times \mathrm{D}_{x}\varphi \cdot h \\[2mm] &= \frac{4 \Vert x \Vert^{2} \left\langle x,h \right\rangle}{1+\Vert x \Vert^{8}} \\ \end{align*} $$

Which gives you the expected result for $\mathrm{D}_{x}f \cdot x$.

pitchounet
  • 6,576
  • Damn it! There's always an easier way and I categorically get stuck in the harder (if not impossible) way... Thank you for the solution. You saved me a ton of trouble. – Luka Horvat Feb 01 '14 at 21:40
  • I know what you mean. Computing the partial derivatives $\frac{\partial f}{\partial x_{i}}$ is not always the easiest way. I guess it depends on the context. – pitchounet Feb 01 '14 at 21:45