find dy/dx when root x + 3 (rooty) =5?

Question

the answer would be 1/9 (1 - (5/ rootx)) but how do we do it. im really confused in the steps..

Answer 1

root x + 3 (rooty) =5
3 rooty = 5-rootx ----------- square two side
9y = 25 - 10rootx + x
y = 1/9(25 - 10rootx +x)
dy/dx = 1/9(0 -5/rootx +1)
dy/dx = 1/9 (1 - (5/ rootx))

Answer 2

root x + 3 (rooty) =5
(rooty) =5/3-1/3rootx
y=25/9+1/9x-10/9rootx

y'=1/9-10/9 /2rootx
y'=1/9-5/ (9 rootx)
y'=1/9{1-5rootx)

Answer 3

√x + 3√y = 5
Differentiate implicitly with respect to x,
1/(2√x) + 3y'/(2√y) = 0
Solve for y',
y' = -√y / (3√x)
= -(5/3 - √x /3) / (3√x)
= (1/9)(1 - 5/√x)

Answer 4

Sorry - just had to comment on the Rooty-Tooty ;-)

Answer 5

sqrt(x) + 3*sqrt(y) = 5

Subtract sqrt(x) from LHS and RHS

3*sqrt(y) = 5 - sqrt(x)

Divide LHS and RHS by 3

sqrt(y) = (1/3)[5 - sqrt(x)]

Differentiate

(1/2)/sqrt(y) (dy/dx) = (1/3)(-1/2)/sqrt(x)

dy/dx = [(1/3)(-1/2)/sqrt(x)]/[(1/2)/sqrt(y)]
dy/dx = [(1/3)(-1/2)/sqrt(x)]*[sqrt(y)/(1/2)]

Recall sqrt(y) = (1/3)[5 - sqrt(x)]

Thus

dy/dx = [(1/3)(-1/2)/sqrt(x)]*[(1/3)(5 - sqrt(x))/(1/2)]
dy/dx = [(1/9)(-1)/sqrt(x)]*[5 - sqrt(x)]
dy/dx = -(1/9) *[5 - sqrt(x)] /sqrt(x)
dy/dx = -(1/9) *[5/sqrt(x) - 1)]
dy/dx = (1/9) *[1 - 5/sqrt(x)]