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MathGroup Archive 1998

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Re: Re: How do u go about doing this---?



Yes, Sorry, Thanks
Jens

> Oops Jens,
> you overlooked the fact that he wanted the sum of squares of the first
> derivative, not the sum of the second derivatives.
> Happens to the best of us, at least some of the time. I goof myself now &
> then, I know the feeling. But *do* go on replying. The least we, non-guru's,
> can do is relieve some of the workload of the "Abbotts & Hintons", letting
> them treat the stuff we can't.
> 
> with sympathy
> 
> wouter.
> 
>    >> Show that
>    >>                 (df/dx)^2 + (df/dy)^2    =
> >Hi, it can't be shown because with
> >f[x,y]->(Sin[x]+Cos[y])/(Sin[x]-Cos[y])
> >
> >one gets for
> >
> >D[f[x,y],{x,2}]+D[f[x,y],{y,2}]=
> >
> >(-4*(Cos[x]^2*Cos[y] + Sin[x]*Sin[y]^2))/(Cos[y] - Sin[x])^3
> >
> >what is clearly different from Your "result"
> >
> >Hope that helps
> >  Jens
> >
> >
> 
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