# Jacobian Determinants (Implicit differentiation)

• Oct 31st 2010, 09:20 AM
jegues
Jacobian Determinants (Implicit differentiation)
The equations,

$\displaystyle x^{2} + y +3s^{2} + s = 2t-1, y^{2} - x^{4} +2st +7 = 6s^{2}t^{2}$

define s and t as functions of x and y. Find $\displaystyle \frac{\partial s}{\partial x}$ 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 $\displaystyle \frac{\partial s}{\partial x}$ 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?
• Oct 31st 2010, 11:26 AM
Opalg
Your answer $\displaystyle \frac{\partial s}{\partial x} = 2x^3$ (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

$\displaystyle x^2+y = 1,\quad y^2-x^4 +7 = 0.$

After factorising $\displaystyle x^4-y^2$ as $\displaystyle (x^2+y)(x^2-y)$ 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 $\displaystyle \frac{\partial s}{\partial x} = 16$, not –16 as the book claims. It looks as though the book is wrong.