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April 19, 2015

April 19, 2015

Posted by **josephine** on Thursday, March 1, 2012 at 10:50pm.

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?

- college implicit HELP! -
**Steve**, Friday, March 2, 2012 at 11:18amDidn'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'

- college implicit HELP! -
**josephine**, Friday, March 2, 2012 at 8:46pmhow u get 2x/√xy? i thought 4x/√xy.

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