THE FIRST FEW CELLS (THAT LOOK EMPTY) CONTAIN THE CODE FOR THE COMMANDS USED BELOW.  I WOULDN'T RECOMMEND DELETING THEM.

iterate[functmm_,startmm_,itmm_,skipmm_]:=MatrixForm[N[NestList[functmm,Nest[functmm,N[startmm],skipmm],itmm],10]]

newtons[functionmm_, varmm_, startmm_, itmm_] :=
(newt[zmm_] :=  zmm - (functionmm /. varmm-> zmm)/(D[functionmm,varmm] /. varmm->zmm);
iterate[newt,startmm, itmm, 0])

iterateexact[functmm_,startmm_,itmm_,skipmm_]:=MatrixForm[NestList[functmm,Nest[functmm,startmm,skipmm],itmm]]

newtonsexact[functionmm_, varmm_, startmm_, itmm_] :=
(newt[zmm_] :=  zmm - (functionmm /. varmm-> zmm)/(D[functionmm,varmm] /. varmm->zmm);
iterateexact[newt,startmm, itmm, 0])

The command newtons uses Newton's method to attempt to find one zero of a function f.  Given an equation of the form f(x)=0, the syntax for newtons is 

newtons[f(variable), variable, starting value, number of iterates].

For example, to find one solution to the equation cos x - x = 0, we would plot the function to find a good estimate for the location of the zero.  (I won't put the graph here to keep the file small.)

Plot[Cos[x]-x,{x,0,2}]

It looks like x=1 would be a good guess.  I would choose 15 iterates, and conclude that one zero is at around 0.739085.

newtons[Cos[x]-x,x,1,15]

newtons[x^2-2,x,1,5]

newtonsexact[x^2-2,x,1,5]

One more example.  Let us find all of the zeroes of the function f(w)=(w^5)-4(w^4)+3w-3.  First, graph it.

Plot[(w^5)-4(w^4)+3w-3,{w,-2,5}]

There appears to be only one zero near w=4.

newtons[(w^5)-4(w^4)+3w-3,w,4,10]

The zero is approximately 3.96399.