Question

Find the slope of the tangent line to the curve below at the point ( 5 , 5 ) .

√3x+2y+√4xy=15

slope=

Answer 1

Your given point does not even lie on the curve, so any calculations would
be futile.
Check your typing

Answer 2

√3x+2y + √4xy =15

√ this goes on for the whole equation for equation for both
√3x+2y √4xy

Answer 3

√(3x)+2y+2√(xy) = 15
3/(2√(3x)) + 2y' + 1/(√(xy)) (y + xy') = 0
2y' + x/√(xy)y' = -3/(2√(3x)) - y/√(xy)

y' = -(3/(2√(3x)) + y/√(xy)) / (2 + x/√(xy))
= -(y√(3x) + 2y√(xy))/(4xy + 2x√(xy))
= -(√(3y) + 2) / √x(4√y + 2x)

Answer 4

so whats the slope

Answer 5

so, at (5,5),
y' = -(√15 + 2) / √5(4√5 + 10)
better check my algebra above

Answer 6

isnt it a whole number the slope?

Answer 7

ahhh, I suspect you meant:
√(3x+2y) + √(4xy) = 15 , strange how a few innocent brackets make the
whole thing come alive.
(3x+2y)^(1/2) + (4xy)^1/2 = 15
(1/2)(3x+2y)^(-1/2) (3 + 2dy/dx) + (1/2)(4xy)^(-1/2)(4xdy/dx + 4y) = 0
(3 + 2dy/dx)/√(3x+2y) + (4xdy/dx + 4y)/√(4xy) = 0
sub in your point (5,5)
(3+2dy/dx)/5 + (20dy/dx + 20)/10 = 0
multiply each term by 10
6 + 4dy/dx + 20dy/dx + 20 = 0
24dy/dx = -26
dy/dx = -13/12

Answer 8

thanks for the help, ill note this formula

Answer 9

I graphed √(3x+2y) + √(4xy) = 15
and the tangent line of y-5 = (-13/12)(x-5) on Desmos and it
verfied. copy and paste the link below

www.desmos.com/calculator