Calculus (Implicit Differentiation)?

Question

I had a test yesterday n not sure if I did this question right:

Q:squareroot(x+y)=1+(x^2)(y^2)

I got the answer y'=[1/2(x+y)^(-1/2)-2x(y^2)]/2(x^2)(y)

(P.S: d/dx both sides and find dy/dx)

Thanks a lot

I used the first suggested method and I got a different answer, 4x(y^2)+4(x^3)(y^4)/1-4y(X^4)-4y(x^2)

Answer 1

Take the derivative of each side
(x+y)^(1/2) = 1+(x^2)(y^2)
(1/2)(x+y)^(-1/2) (1+y') = 2x(y^2) + (x^2)(2y y')
(1/2)(x+y)^(-1/2) + (1/2)(y')(x+y)^(-1/2) = 2xy^2 + 2x^2y y'
(1/2)(y')(x+y)^(-1/2) - 2x^2y y' = 2xy^2 - (1/2)(x+y)^(-1/2)
y' ((1/2)(x+y)^(-1/2) - 2x^2y) = 2xy^2 - (1/2)(x+y)^(-1/2)
y' = [2xy^2 - (1/2)(x + y)^(-1/2)]/[(1/2)(x + y)^(-1/2) - 2x^2y]

Answer 2

I think your final solution got chopped off, but it looks as though you were on the right track.

I just worked it and got

dy/dx=[2xy^2-(1/2)(x+y)^(-1/2)] all over

[(1/2)(x+y)^(-1/2)-2x^2y]

Answer 3

Try to do it this way:
First apply the square on both the sides to get rid of the square root.
Then use implicit differentiation.
Collect dy/dx on one side and get the final answer.

Answer 4

i have self belief you probably did it with out using implicit differentiation. y² = ln x Doing it with implicit differentiation: 2y dy/dx = one million/x dy/dx = one million/(2xy) evaluate at x = e, y = one million dy/dx = one million/(2(e)(one million)) = one million/(2e) excellent answer, yet your differentiation is using chain rule.