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Re: Inverse Function

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  • Subject: [mg121472] Re: Inverse Function
  • From: Daniel Lichtblau <danl at>
  • Date: Fri, 16 Sep 2011 05:48:42 -0400 (EDT)
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On 09/15/2011 03:40 AM, Giuseppe Benedetti wrote:
> Hello to everyone,
> I have a function of two variables g[k,s] and I want to compute (symbolically) its inverse with respect to the first variable k (ie keeping s constant). I am therefore looking for a function G[g,s], which I then need to differentiate (twice) with respect to the second variable s. Does anybody know how to do it???
> Thank you in advance.
> Giuseppe.

Start with the identity

G[g[k, s], s] - k == 0

Differentiate once and twice with respect to k and likewise with respect 
to x. All these derivatives are zero. Use that to solve for D[0,2][G] in 
terms of derivatives of g.

In[21]:= Derivative[0, 2][G][g[k, s], s] /.
  Solve[{D[G[g[k, s], s] - k, k] == 0,
    D[G[g[k, s], s] - k, {k, 2}] == 0, D[G[g[k, s], s] - k, s] == 0,
    D[G[g[k, s], s] - k, {s, 2}] == 0}, {Derivative[2, 0][G][g[k, s],
     s], Derivative[0, 2][G][g[k, s], s],
    Derivative[1, 0][G][g[k, s], s], Derivative[0, 1][G][g[k, s], s]}]

Out[21]= {-2*Derivative[0, 1][g][k, s]*
    Derivative[1, 1][G][g[k, s], s] -
      (Derivative[0, 2][g][k, s] - (Derivative[0, 1][g][k, s]^2*
                Derivative[2, 0][g][k, s])/Derivative[1, 0][g][k, s]^2)/
        Derivative[1, 0][g][k, s]}

In principle I would have thought we'd also need the second order mixed 
derivative, but it seems to have worked without that.

Daniel Lichtblau
Wolfram Research

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