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A function $f:R^n \to R$ belong to $C^2$. So $Df$ will be a matrix of order "1 by n". So now the range set of function $Df$ will be all row matrix of order "1 by n" while domain set will remain same(set of all column matrix of order "n by 1"). Then how can I handle this function? How would I determine $D^2f$? Can anyone please help me out?

cmi
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    Have you ever done it for some $f:\mathbb R^2 \to \mathbb R$? In any case, see the hessian matrix – Andres Mejia Jul 24 '18 at 15:28
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    for functions $f:\mathbb R^m \to \mathbb R^n$ you will probably need the concept of tensors – Andres Mejia Jul 24 '18 at 15:30
  • I have just started reading MVC. Can you please tell me about the function $Df$? what is it's domain and range? How would I differentiate it again?@AndresMejia – cmi Jul 24 '18 at 15:32
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    $Df$ is a "linear approximation" of a function $\mathbb R^n \to \mathbb R$. Try to compute if for $f(x,y)=x+y$. This is the "trivial case" where the function is already linear. Try then to compute $f(1,1)+Df(1,1)(\mathbf{x}-(1,1))$. Next you can try a function $f(x)=x^2+y^2$ and do the same computation. Evaluate the difference between $f$ and the expression I give for $\mathbf{x}=(1,1.1)$ and $\mathbf x = (1.1,1)$. I'm not really a teacher, so I don't know the best way to internalize this kind of thing. Try here or Hubbard's bk. – Andres Mejia Jul 24 '18 at 15:40
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    There is the following property of the standard derivative: $f(a+\Delta x) \approx f(a)+f^{\prime}(a) \cdot (\Delta x)$. In this sense, $f^{\prime}(a)$ is the best linear approximation near $a$ (since a one dimensional matrix is multiplication. The situation is similar in higher dimensions, but linear transformations now take the form of matrices. In your case $n \times 1$ matrices applied to $n$-vectors – Andres Mejia Jul 24 '18 at 15:43
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    The trouble with the second derivative is that in each co-ordinate direction, you now have to see how an $n \times 1$ "row" is behaving, so for each $x_i$ you get a column, and hence at the end an $n \times n$ matrix. – Andres Mejia Jul 24 '18 at 15:45
  • I am trying to get your point. Once I am done , I will tell how much I have understood.@AndresMejia – cmi Jul 24 '18 at 15:47
  • Don't look too closely! They're just kind of rambles. Try here or here for much better discussions. – Andres Mejia Jul 24 '18 at 15:51
  • If $f(x,y) = x+y$ then $$Df : \begin{bmatrix} X_1 \ \ X_2 \end{bmatrix} \to (1 , 1)$$ Now how can I differentiate $Df$?@AndresMejia – cmi Jul 24 '18 at 16:03
  • Well $Df$ IS constant so I think there is a pretty fair guess for the derivative. – Andres Mejia Jul 24 '18 at 16:05
  • But I am feeling inconvenient to handle because it's domain point is a column matrix where it's image is a row matrix.@AndresMejia – cmi Jul 24 '18 at 16:07
  • How can I determine $D^2f$ from the definition. Can I use $lim \frac {f(a+h) - f(a) - B.h}{|h|}$ this definition to determine it's derivative? Here $f(a+h)$ , $f(a)$ are row matrices of order "1 by 2" where $h$ is a column matrix of order "2 by 1" Then how can I get a Matrix $B$?@AndresMejia – cmi Jul 24 '18 at 16:11
  • I'm not sure how much more commenting is SE-acceptable, but the definition can be used to show that you have found the derivative but is in practice not helpful in determining what the derivative is. You should instead calculate the jacobian and then prove that this is indeed the derivative. – Andres Mejia Jul 24 '18 at 16:16
  • How can I determine $D^2f$ from the definition. Can I use $lim \frac {f(a+h) - f(a) - B.h}{|h|}$ this definition to determine it's derivative? Here $f(a+h)$ , $f(a)$ are row matrices of order "1 by 2" where $h$ is a column matrix of order "2 by 1" Then how can I get a Matrix $B$ because $Bh$ will always be a column matrix?@AndresMejia – cmi Jul 24 '18 at 16:16
  • okay. I am using definition because this a petty problem. But there must be a problem . Can you please point out where did I go wrong?@AndresMejia – cmi Jul 24 '18 at 16:18

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