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Let $$ f(x_1,x_2) = \frac{1}{\sqrt{x_1^2+x_2^2+1}} \left(x_1,x_2,1 \right) $$

For $i = 1,2$ we have

$$ \partial_{x_i} f = \frac{1}{\sqrt{x_1^2+x_2^2+1}}\left(\partial_{x_i}x_1,\partial_{x_i}x_2,0 \right) +\frac{-x_i}{\left(x_1^2+x_2^2+1\right)^{3/2}}(x_1,x_2,1) = \frac{1}{\left(x_1^2+x_2^2+1\right)^{3/2}} \left[\left(x_1^2+x_2^2+1\right)\left(\partial_{x_i}x_1,\partial_{x_i}x_2,0\right) - x_i(x_1,x_2,1) \right] $$

The gradient of such function, after some computation I get the following matrix

$$ \nabla f = \frac{1}{\left(x_1^2+x_2^2+1\right)^{3/2}} \begin{pmatrix} x_2^2+1 & - x_1x_2 & -x_1 \\ -x_1x_2 & x_1^2+1 & -x_2 \end{pmatrix} $$

Is it correct?

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  • $\begingroup$ the gradient is not a matrix. Maybe you mean the jacobian matrix. $\endgroup$
    – Masacroso
    Feb 28, 2018 at 15:53
  • $\begingroup$ The gradient is a matrix if your function is a vector, which is my case. $\endgroup$
    – user8469759
    Feb 28, 2018 at 15:54
  • $\begingroup$ @Masacroso: tha gradient of a vector function is an standard definition in some contexts. (eres el mismo del rincon?) $\endgroup$
    – yemino
    Feb 28, 2018 at 16:03
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    $\begingroup$ @Masacroso It's not my language... math.stackexchange.com/questions/156880/…, this guy calls "gradient" essentially the same thing. $\endgroup$
    – user8469759
    Feb 28, 2018 at 16:09
  • $\begingroup$ @user8469759 alright, first time I heard of it $\endgroup$
    – Masacroso
    Feb 28, 2018 at 16:12

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The calculus of the partial derivates are correct. But the definition that I know of the matrix gradient is the transpose matrix that you wrote.

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  • $\begingroup$ I think that depends from how you consider vectors, right? (If rows or columns) $\endgroup$
    – user8469759
    Feb 28, 2018 at 15:55
  • $\begingroup$ In definition that I know, the gradient of a vector function $f$ is a matrix which rows are the gradient of the components of $f$, so the gradient is the same matrix if $f$ is a row or a column vector. But you must to revise the definition that you are using. $\endgroup$
    – yemino
    Feb 28, 2018 at 16:02
  • $\begingroup$ Actually I'm not using any definition. I'll check later my analysis book. $\endgroup$
    – user8469759
    Feb 28, 2018 at 16:04
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    $\begingroup$ @yemino I saw sometimes multivariable calculus written as $xA$ instead of $Ax$, for some jacobian matrix $A$. It is rare (as rare as call "gradient" to the jacobian matrix :p) but exists. Im the same guy of the rincon. $\endgroup$
    – Masacroso
    Feb 28, 2018 at 16:16

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