If x (r, th) =r cos (r,th) and y (r,th) = r sin(th). The Jacobian of the transformation is

[ [ cos(th), -rsin(th)], [ sin(th), r cos(th) ]]. Given a function f(x,y) of x and y, in new coordinates r, th the value of  fr  = fx cos(th)  + fy sin(th), fth =fx .(-r sin(th))  + fy r cos(th).

Now, frr = ( fx cos(th) + fy sin(th))x  cos(th) +( fx cos(th) + fy sin(th))y sin(th)

              = fxx cos2 (th) + fyx sin(th).cos(th) + fxy cos(th)sin(th) + fyy sin2 (th) 

                  Assuming fxy = fyx, then this can be written as

               =  fxx cos2(th) + 2fxy cos(th)sin(th) + fyy sin2 (th) 

If you know how to express first order derivatives in these coordinates, you just use this sequentially. In this way, for instance the Laplacian in curvilinear coordinates is found by calculating the divergence of a gradient. See wikipedia (link below) and  Laplacian in curvilinear coordinates (e.g., see 2nd link below). 

http://en.wikipedia.org/wiki/Curvilinear_coordinates

http://www.physics.uoguelph.ca/~jhd/phys2460/Week4/L11_eqns.pdf

If x (r, th) =r cos (r,th) and y (r,th) = r sin(th). The Jacobian of the transformation is

[ [ cos(th), -rsin(th)], [ sin(th), r cos(th) ]]. Given a function f(x,y) of x and y, in new coordinates r, th the value of  fr  = fx cos(th)  + fy sin(th), fth =fx .(-r sin(th))  + fy r cos(th).

Now, frr = ( fx cos(th) + fy sin(th))x  cos(th) +( fx cos(th) + fy sin(th))y sin(th)

              = fxx cos2 (th) + fyx sin(th).cos(th) + fxy cos(th)sin(th) + fyy sin2 (th) 

                  Assuming fxy = fyx, then this can be written as

               =  fxx cos2(th) + 2fxy cos(th)sin(th) + fyy sin2 (th) 

Thank you Sir for the reply

This is my problem.

If (x,y)=f1 (u,v) and N=f2 (x,y)

Then {∂N/∂x; ∂N/∂y}=J-1 {∂N/∂u); ∂N/∂v}

Here J=[∂x/∂u ∂y/∂u

             ∂x/∂v ∂N/∂v];

Similarly I need to calculate the relation between {∂2 N/∂x2 ; ∂2 N/∂y2 } and {∂2 N/∂u2 ; ∂2 N/∂v2 } something like {∂2 N/∂x2 ; ∂2 N/∂y2 } = [a matrix]* {∂2 N/∂u2 ; ∂2 N/∂v2 } . Is this possible? Or how to evaluate this? Please be more specific.

So    [∂N/∂u, ∂N/∂v]t   = J [∂N/∂x,  ∂N/∂y ]t  or operators   [∂/∂u, ∂/∂v] t = J [∂/∂x,  ∂/∂y ]t . 

    or  [∂N/∂u, ∂N/∂v]t =  [∂N/∂x.∂x/∂u + ∂N/∂y . ∂y/∂u, ∂N/∂x.∂x/∂v + ∂N/∂y . ∂y/∂v ]t ,     Therefore   [ ∂ 2N/∂u2 ,  ∂2 N/∂v2 ]t  = { J [∂/∂x, ∂/∂y ]t }2  [ N, N ]t

= J [∂/∂x, ∂/∂y ]t J [∂N/∂x, ∂N/∂y ]t

Sir , could you please give me some references..texts or articles?

Answer to your original question is , you may see as given above. For reference you may see a good book on differential calculus or Schaum Series. If you want form for            [ ∂ 2N/∂x2, ∂2N/∂y2 ]t = J-1[∂/∂u), ∂/∂v ]t J-1 [∂N/∂u),∂N/∂v ]t .

Thanks a lot.

I agree with Prof. Mittal

I had the same question, and I found this equation.

I think this could be helpful

Regards