Implicit Differentiation for Tangent Line Equations

[EvilBunny]

Homework Statement

I need to find a tangent line equation but my main problem is isolating actually.

sqrt ( 2x + 2y ) + sqrt ( 3xy) = 13

The Attempt at a Solution

0.5 ( 2x + 2y ) ^ -0.5 (2+2y') + 0.5(3xy) ^ -0.5 ( 3 ( y + y' x ) ) = 0

this is where I get stuck

 

[Martin III]

The differentiation looks good to me. Now you just need to simplify.

If you're having trouble, I'd multiply out
0.5 ( 2x + 2y ) ^ -0.5 (2+2y')

and
0.5(3xy) ^ -0.5 ( 3 ( y + y' x ) )

and then pull a y' out of two terms, move the remaining terms to the other side and divide by what's multiplied with y'.

 

[HallsofIvy]

Don't let the fact that it looks complicated stop you. The equation you get by "implicit differentiation" is always linear in y'. Separating y' as Martin III said should be easy.

0.5(2x+2y)^1/2(2x+ 2y')+ 0.5(3xy)^1/2(3y+ 3y')= 0
(2x+2y)^1/2x+ (2x+2y)^1/2y'+ (3/2)y(3xy)^1/2+(3/2)y'(3xy)^1/2= 0
[(2x+2y)^1/2+ (3/2)(3xy)^1/2]y'= -(2x+2y)^1/2x- (3/2)y(3xy)^1/2.

In fact, since you say you are to "find a tangent line equation", you must be given some point at which to find the tangent line. If you just put the values you are given for x and y into the original equation you should find it much easier.