Question

How do you differentiate `y=cosy/x-x^2y`?

Answer 1

`dy/dx=-(cosy+2x^3y)/(x(x^3+x+siny))`

Explanation:

`d/dx[y=cosy/x-x^2y]`

`dy/dx=(dy/dx(-siny)*x-cosy)/x^2-2xy-x^2dy/dx`

`dy/dx+x^2dy/dx=(dy/dx(-xsiny)-cosy-2x^3y)/x^2`

`x^2(dy/dx+x^2dy/dx)=dy/dx(-xsiny)-cosy-2x^3y`

`dy/dx(x^4+x^2+xsiny)=-cosy-2x^3y`

`dy/dx=-(cosy+2x^3y)/(x(x^3+x+siny))`