college implicit HELP!
posted by josephine .
I WANNA KNOW I'M DOING RIGHT.
Find the slope of the tangent line to the curve √(3x+2y) + √(4xy)=11.6 at the point (6,4).
THAT WHAT I DID: (3x+2y)^1/2+(4xy)1/2=11.6
next step: 1/2(3x+2y)^1/2*y'+1/2(4xy)*y=0
y'= (3x+2y)/(4xy) right?

Didn't apply the chain rule completely:
(3x+2y)^1/2+(4xy)1/2=11.6
(1/2)(3x+2y)^1/2 *(3+2y') +(1/2)(4xy)^1/2 * 4(y + xy') = 0
y'(2y/√(3x+2y) + 2x/√xy) = (3/√(3x+2y) + 2y/√xy)
3/√(3x+2y) + 2y/√xy
 = y'
2y/√(3x+2y) + 2x/√xy
you can massage that more if you wish. The key is to note that
d/dx(3x+2y) = 2 + 2y'
and
d/dx(4xy) = 4y + 4xy' 
how u get 2x/√xy? i thought 4x/√xy.