No, taking the sqrt of both sides gets you y = +/- the sqrt of the stuff on the right side. So you're have to do two problems of regular differentiation.
Only substitute y in for what y equals if you used logarithmic differentiation. For example, if you start of with y = x^x, you will use log differentiation which is taking ln's of both sides and doing implicit differentiation, but at the end you're supposed to put what y equals. But not if you started with a curve you can't solve for. Like if you get e^xy = sin(x/y) + (y^87) (x^100), I'm pretty sure you can't solve for y lol.
Another way to implictly differentiate equations with two variables is something you learn in multivariable calculus. Manipulate your equation so that you get everything = 0. So in the first example in the thread, we want y^2 -((x^2 - 9)/(x^2 +9)) = 0. Let f(x,y) = y^2 -((x^2 - 9)/(x^2 +9)). Now find -Fx/Fy, which are partial derivaties where with Fx, you just pretend y is a constant and differentiate with respect to x, and do the opposite with Fy, and you get the same thing you would if you did it the calc1 method. | 
11-16-2007, 08:11 PM
