Thread: Jacobian Determinants (Implicit differentiation)

The equations,



define s and t as functions of x and y. Find when s = 0 and t = 1. Assume x > 0.

I've gone through the whole process defining the functions F and G and using Jacobian determinants to solve for but I'm still stuck with some x's in my answer that I can't figure out eliminate.

The answer listed is -16, but I after all my computation I came up with 2x^3.

You can see that if you define the first equation as a function F(s,t,x,y) = 0, and G(s,t,x,y) = 0 the partial with respect to x of equation G (-4x^3) won't go away, even after evaluating s and t at 0 and 1 repsectively.

Any ideas/suggestions?

Your answer (at the point where (s,t) = (0,1)) is correct. To find the value of x, just put s=0 and t=1 in the original equations. You get



After factorising as and using the given information that x>0, you should be able to solve those equations to get x=2. That gives the answer as , not –16 as the book claims. It looks as though the book is wrong.