I've just studied the implicit function theorem and if we assume the theorem is true then we can easily compute the following:(adsbygoogle = window.adsbygoogle || []).push({});

id_n = D(id_n)_x = D({f^-1}°{f})_x = {D(f^-1)_f(x)} ° {Df_x}

where D(*)_a means the derivative of * at x.

OKay.... so this was very straightforward until I began to think about D(id_n)_x.

I know that this id_n is not the same as the first id_n. The id_n of D(id_n)_x is the matrix representing the transformation induced by the identity function. i.e. f(x_1,x_2,...x_n) = (x_1,.... x_n). But then I began to wonder why it says "at x" (i.e. D(id_n)_x).

In the one dimensional case, f(x) = x. Or another way to look at it f(x) = 1°x. Where ° means composition and is ordinary multiplication, here. I think my problem is that I'm confusing the function with its "action" (by that I mean, I think "f=1" but f(x)=x). This is a very subtle distinction and so I hope that someone that appreciates this can give a not so off hand reply (i.e. are we differentiating the transformation of basis (identity) or the function? what does it look like?)

Anyway, so I'm doing to assume that in one dimension the first equation reduces to:

1=D(1)_x = .....

But I'm not sure what it means to say D(1)_x

Also could it be that the notes I'm reading are treating "x" as a "point" rather than a variable (and that the notation D(*)_x avoids explicitly stating w.r.t. which variable, other than "all of them/total derivative").

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# Derivative of transformation of basis in R^n

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