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college implicit HELP!

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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?

  • college implicit HELP! -

    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'

  • college implicit HELP! -

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

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