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.

To find the slope of the tangent line to the curve at the point (6,4), you correctly took the equation and found the derivative (y') in respect to x.

Let's go through the process step by step:

1. Start with the equation: √(3x+2y) + √(4xy) = 11.6.

2. To find the derivative, you need to differentiate both sides of the equation with respect to x.

Differentiating the left side:
The derivative of √(3x + 2y) with respect to x is 1/2(3x + 2y)^(-1/2) multiplied by the derivative of (3x + 2y) with respect to x, which is 3.

Differentiating the right side:
The derivative of 11.6 with respect to x is 0 since it is a constant.

3. Bring the derivative terms together and simplify:
1/2(3x + 2y)^(-1/2) * (3) + 0 = 0

Simplifying further:
3/2(3x + 2y)^(-1/2) = 0

4. Solve for y':
To isolate y', multiply both sides by 2(3x + 2y)^(1/2):
3 = 0

Therefore, we have:
3 = 0

In this case, we have a contradiction, indicating that there is no solution for y'. The curve does not have a well-defined tangent line at the point (6,4). It may be a singular point or have a different behavior at that specific point.