does anyone know how to solve this calculus question!?

Question

Suppose variables x and y are related by: xy + 2 sin (y^2+1) = sin (x^2y^2). Find an expression, in terms of both x and y , for the derivative y'(x).

Answer 1

xy + 2sin(y² +1) = sin(x²y²)

Using implicit differentiation

y + xy' + 2cos(y² +1).2yy' = cos(x²y²).(2xy² + x².2yy')

Thus xy' + 4yy'cos(y² +1). - 2x²yy'cos(x²y²) = 2xy²cos(x²y²) - y

So y'[x + 4ycos(y² +1). - 2x²ycos(x²y²)] = y[2xycos(x²y²) - 1]

Whence y' = y[2xycos(x²y²) - 1] / [x + 4ycos(y² +1). - 2x²ycos(x²y²)]

Answer 2

implicit differentiation

xy+2sin(y^2+1)=sin(x^2y^2)

y+x(dy/dx) + 2cos(y^2+1)2y(dy/dx)= cos(x^2y^2)(2xy^2+2x^2y(dy/dx))

I don't have time to factor now, but treat (dy/dx) as a variable and put all of the terms with that variable on the same side, and put the rest on the other side of the equation. Then factor out the dy/dx on the side that you put it on. Then divide all the terms that used to have the dy/dx on them over to the other side and you will have dy/dx by itself and solved for

Answer 3

pie=X,y?