find dy/dx for cos(x^2y^2)/sin(x^2y^2)=2x

Question

The original equation is: cos(x^2y^2)/sin(x^2y^2)=2x

 

I simplified it to cot(x^2y^2)=2

 

I used chain rule to get -csc^2u * du/dx and found du/dx= (2x^2yy'+2y^2x)

new equation: -csc^2(x^2y^2)(2x^2yy'+2y^2x)=2

 

I'm not sure if I've done it correctly so far, and i'm not sure where to go from here. 

Answer 1

Replacing cos/sin with cot probably makes things easier,

d/dx[ cot(x²y²) = 2x ]

 

-csc²(x²y²) * (2xy² + 2x²yy') = 2

 

You want to solve for y'. I would first factor out common factors before trying to isolate the y'.

 

-2xy csc²(x²y²) (y + xy') = 2

 

y + xy' = -1 / [ xy csc²(x²y²) ]

xy' = -1 / [ xy csc²(x²y²) ] - y

y' = -1 / [ x²y csc²(x²y²) ] - y/x

 

If you want to write as one fraction:

 

y' = [-1 + xy² csc²(x²y²)] / [ x²y csc²(x²y²) ]