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 > > > > > > > > > >