You can usually lose all the trig functions with cos u -> (1 - u^2)/(1 + u^2), sin u -> 2 u / (1 + u^2), and similar for v. The new u has range [-1,1], assuming the original u was real. This is the ancient tan(u/2) -> u substitution. In your case, you also want sin((u-v)/2) to be a rational function, so u & v should be halved: cos(u/2) -> (1-u^2)/(1+u^2) etc. cos(u) -> (1 - 6 u^2 + u^4)/(1+u^2)^2 sin(u) -> (4 u - 4 u^3)/(1+u^2)^2 sin((u-v)/2) = sin(u/2 - v/2) = sin(u/2) cos(v/2) - sin(v/2) cos(u/2) -> [ 2u(1-v^2) - 2v(1-u^2) ] / [(1+u^2)(1+v^2)] The sqrts will still mess up the derivatives. If you are looking for critical points, you can multiply through by (1+u^2)(1+v^2) before differentiating to simplify things a little. Rich -------- Quoting Marc LeBrun <mlb@well.com>:
="Dan Asimov" <dasimov@earthlink.net> f(u,v) = sqrt(5+4cos(u)) + 2.5*sin((u-v)/2) + sqrt(11-6cos(v)-2sin(v)).
Also, computer or not, it might helpful to get rid of the trig here by first substituting s:=sin u, t:=sin v (or something similar) leaving mere algebra.
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