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Let us assume that you simply have a point: $(x_1, x_2).$ You also have a transformation $H,$ that maps this single point to a new point: $(y_1, y_2).$ So $y_1 = h_1(x_1, x_2),$ and $y_2 = h_2(x_1, x_2).$

My question is simple:

Given a new point of $(x_1 + dx_1, x_2),$ what is the new transformed point equal to, in terms of any partials necessary? (aka, partials of $h_1$ w.r.t $x_1,$ or $x_2,$ and $h_2$ w.r.t $x_1$ and $x_2$).

EDIT:

$H$ is differentiable.

So my question is, let us get the first transformed point, $y_1$ in this case. This is equal to $h_1(x_1 + dx_1, x_2)$. What is this actually equal to, in terms of partials? Similarly, get the transformed point $y_2$, in this case will be $h_2(x_1 + dx_1, x_2)$. What is this in terms of the partials?


Edit 2: I am trying to understand bottom part of page 2 of this paper PDF. To be precise, I am trying to understand how the author got $(x_1+dx_1,x_2)→(y_1+a,y_2+b).$ I think it should be $(x_1+dx_1,x_2)→(y_1+a,y_2+c).$ (Page two, towards the bottom).

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1  
What assumptions are you making on $H$? You seem to want it to be differentiable, and you also seem to be asking what its partial derivative with respect to $x_1$ is. Well, it's... its partial derivative with respect to $x_1$. I don't know what else to say. –  Qiaochu Yuan Jul 19 '12 at 20:22
    
Are you asking for something like $H(x_1 + dx_1, x_2) = H(x_1,x_2) + \int_0^1 \partial_s H(x_1 + s dx_1, x_2) ds = H(x_1, x_2) + \int_0^1 D(H)(x_1 + s dx_1, x_2) * dx_1 ds$? –  Cocopuffs Jul 19 '12 at 20:41
    
@QiaochuYuan Please see my edit. Yes its differentiable. –  Mohammad Jul 19 '12 at 20:42
    
@Cocopuffs I think so yes. I am trying to understand bottom part of page 2 of this paper. –  Mohammad Jul 19 '12 at 20:43
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@Mohammad You're right. –  Cocopuffs Jul 19 '12 at 21:01

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