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3

Square brackets in Mma are used only to specify function arguments.Your function is equivalent to the following (by using FullSimplify[ ]) f[x_, y_, z_] := -((x (-1 + x + y) + (x + y) z)/ Sqrt[(-1 + x + y) (x + y) (-1 + x + z) (x + z)]) and then {D[f[x, y, z], x] , D[f[x, y, z], y] , D[f[x, y, z], z]}


3

At first glance it seems hopeless - you have all those hyperbolic trig and regular trig functions in your exponent. But then you notice that that is all superfluous since the only two variables you care about, X2 and P2 don't go into any of those functions, so that other stuff is just a distraction. {X1,P1,r, φ, θ, η} are all just constants! What you are ...


2

Besides getting the syntax as discussed by belisarius, FullSimplify will be very helpful to you in this case. f[x_, y_, z_] := Sqrt[(x + y) (1 - x - y) (x + z) (1 - x - z)] * (x/((x + y) (x + z)) - y/((x + y) (1 - x - z)) - z/((1 - x - y) (x + z)) + (1 - x - y - z)/((1 - x - y) (1 - x - z))) D[f[x, y, z], x] // FullSimplify ((-1 + 2*x + y + ...


1

When I first made this answer I was bleary eyed and didn't realize that Willinski's answer matched the question except that he made a nice edit by replacing, for example, 1.5 with 3/2. This answer is in addition to Willinski's fine work. I followed his procedure. I wanted to do a numerical study and try to find the region of interest. A = 1; J = 1; c = 1; ...


1

I think this is impossible, i.e there's no way to know if an arbitrary function is continuous unless you are given some additional information about the function. Otherwise a function is just a black box. If you function is made up of only known functions, you could check for Piecewise and check at the interface between definitions for instance, but this ...



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