(* Here is F(x), from #42, p. 154: *)
F[x_]:=Sqrt[x+Sqrt[x+Sqrt[x]]];

(* Now we're going to turn it into a long strange trip: *)
g[x_]:=Sqrt[x];
h[x_]:=x+x^2;
i[x_]:=Sqrt[x];
j[x_]:=x+(2x^2+1-Sqrt[4x^2+1])/2;
k[x_]:=Sqrt[x];
(* So here is F, the composition of all of those: *)
f[x_]:=k[j[i[h[g[x]]]]]

(* This is what I solved to get the j-function: remember the root of that quadratic? Here it is: *)
Solve[y==Sqrt[x+Sqrt[x]],x]

(* Check that they evaluate to the same thing: *)
f[2]
F[2]
N[f[2]]
N[F[2]]
N[{g[2],h[g[2]],i[h[g[2]]],j[i[h[g[2]]]],k[j[i[h[g[2]]]]]}]
N[{Sqrt[2],2+Sqrt[2],Sqrt[2+Sqrt[2]],2+Sqrt[2+Sqrt[2]],Sqrt[2+Sqrt[2+Sqrt[2]]]}]

(* Now check out these derivatives: 
Mathematica handles the first two, and then 
we use the magnificent chain rule to handle the last one: *)
F'[x]
f'[x]
hand[x_]:=(1+(1+1/(2Sqrt[x]))/(2Sqrt[x+Sqrt[x]]))/(2f[x])
k'[j[i[h[g[x]]]]]*j'[i[h[g[x]]]]i'[h[g[x]]]h'[g[x]]g'[x]

(* Graph the three together, just to make sure that they're equal: *)
Plot[{
F'[x],
f'[x],
hand[x],
k'[j[i[h[g[x]]]]]*j'[i[h[g[x]]]]i'[h[g[x]]]h'[g[x]]g'[x]},{x,0,5},AxesOrigin->{0,0}
]

(* Now I'll add a constant to each, so that we can acually see that they're all there: *)
Plot[{F'[x],
1+f'[x],
2+hand[x],
3+k'[j[i[h[g[x]]]]]*j'[i[h[g[x]]]]i'[h[g[x]]]h'[g[x]]g'[x]},{x,0,5},AxesOrigin->{0,0}]