find dy/dx #2?

Question

Find dy/dx?
im new here and will be sure to award someone the best answer for a good response. have a few problems im stuck on.

-- as asked, i'm breaking up my previous post
-- not allowed to use a grapher with these

e^x^2+y^2 = sin (x^2 + y^2)
[thats e to the power of x squared plus y squared]

Answer 1

e^x² + y² = sin (x²+y²)

Differentiating implicitly:

2xe^x² + 2y dy/dx = cos (x²+y²) (2x + 2y dy/dx)

Expanding the product on the right:

2xe^x² + 2y dy/dx = 2x cos (x²+y²) + 2y dy/dx cos (x²+y²)

Collecting like terms:

2y dy/dx - 2y cos (x²+y²) dy/dx = 2x cos (x²+y²) - 2xe^x²

Factoring:

2y (1-cos (x²+y²)) dy/dx = 2x (cos (x²+y²) - e^x²)

Dividing:

dy/dx = x/y (cos (x²+y²) - e^x²)/(1-cos (x²+y²))

Answer 2

well, the first thing to do is to pull everything to one side and set the equation to zero , having done that you will get a function F with the independent variable being x and the dependent variabe y , so you will have F(x, y(x))=0 . using the chain rule of multivariable calculus you get : dF/dx dx/dx + dF/dy dy/dx = 0 , and since dx/dx =1 you will have : dF/dx + dF/dy dy/dx =0 , so solving this for dy/dx you get dy/dx = - dF/dx / dF/dy , and from here it should be easy to find dy/dx because you already have your function as F(x, y(x))= e^x^2 +y^2 - sin(x^2+y^2)=0 , so just take partial derivatives!

Answer 3

1) Let U = x^2 + y^2

2) I think the input equation is:
e^U = sinU

3) Take the derivative of both sides w.r.t. x and let U' = dU/dx
(e^U)U' = (cosU)U';

4) Get U' = dU/dx
U' = 2x + 2yy' where y' = dy/dx

5) Substitute 5) into 4) and solve for y'
y' = -( x / y )

Answer 4

e^(x^2+y^2) (2x+2yy') = cos(x^2+y^2) *(2x + 2yy')
so 2x + 2yy' = 0
so y' = -x/y