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Re: differentiate a function of a function
*To*: mathgroup at smc.vnet.net
*Subject*: [mg75118] Re: [mg75071] differentiate a function of a function
*From*: Bob Hanlon <hanlonr at cox.net>
*Date*: Wed, 18 Apr 2007 04:53:09 -0400 (EDT)
*Reply-to*: hanlonr at cox.net
g[x_,y_]:=Cos[t[x,y]+a[x,y]] Tan[b[x,y]];
dgx=FullSimplify[D[g[x,y],x]/.Tan[b[x,y]]:>k/Cos[t[x,y]]]
Cos[a[x, y] + t[x, y]]*Sec[b[x, y]]^2*Derivative[1, 0][b][x,
y] - k*Sec[t[x, y]]*Sin[a[x, y] + t[x, y]]*
(Derivative[1, 0][a][x, y] + Derivative[1, 0][t][x, y])
Bob Hanlon
---- kem <kemelmi at gmail.com> wrote:
> Thanks a lot!
>
> As a followup, if my function is g[x_,y_]:=Cos[t[x,y]+a[x,y]] Tan[b[x,y]]
> and I take in the same way derivative TrigExpand[D[g[x,y],x]]
> I get complicated expression that starts with
>
> -Cos[t[x,y]] Sin[a[x,y]] Tan[b[x,y]] Derivative[1,0][a][x,y]-Cos[a[x,y]]
> Sin[t[x,y]] Tan[b[x,y]] Derivative[1,0][a][x,y]+...
>
> lets say that i know that
> k = Cos[t[x,y]] Tan[b[x,y]]
>
> how can i use this to simplify my result by substituting 'k' inside
> like instead of every Cos[t[x,y]] Tan[b[x,y]] i put k ?
>
> Thanks
> kem
>
>
> On 4/16/07, Bob Hanlon <hanlonr at cox.net> wrote:
> >
> > f[x_,y_]:=Tan[t[x,y]+a[x,y]];
> >
> > D[f[x,y],x]
> >
> > Sec[a[x, y] + t[x, y]]^2*
> > (Derivative[1, 0][a][x, y] +
> > Derivative[1, 0][t][x, y])
> >
> >
> > Bob Hanlon
> >
> > ---- kem <kemelmi at gmail.com> wrote:
> > > Hi,
> > >
> > > I was wondering how do I define a function in mathematica to be able
> > > to differentiate it etc, where some of the parameters of this function
> > > should be also a function. For example I want to be able to do the
> > > following operations:
> > >
> > > 1) to say that f is a function: f(x,y) = Tan[t(x,y)+a(x,y)]
> > >
> > > 2) take D[f,x] , such that also t(x,y) and a(x,y) will be also
> > > differentiated
> > >
> > > 3) be able to substitute these into some equation like: f_x f_xy = 8
> > >
> > > Thanks a lot
> > >
> > > kem
> > >
> > >
> >
> >
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