# college implicit HELP!

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?

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1. 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'

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posted by Steve
2. how u get 2x/√xy? i thought 4x/√xy.

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