SOLUTION: Find dy/dx for x^3+2xy^2 = sin y

Starting equation is

    x^3 + 2xy^2 = sin(y).    (1)


Here by default we consider x as an independent variable and y as a function of x 

    y = y(x).


Differentiate the given equation (1) (both sides separately).  
Follow standard rules of differentiating. You will get


    3x^2 + 2y^2 + 4xy*y' = cos(y)*y'    (2)


where y' = y'(x) =  for brevity.


Collect and combine the terms in formula (2) in order for to have y' 
as a factor on one side of the equation


    3x^2 + 2y^2 = cos(y)*y' - 4xy*y' ,

    3x^2 + 2y^2 = (cos(y) - 4xy)*y' .


Now express y'


    y' = .    ANSWER


It is the formula which you want to get.

An alternate method is to use partial derivatives of multivariable calculus. 

For f(x,y)=0, we have the chain rule ∂f/∂x = -(∂f/∂y)(dy/dx) 

 -(∂f/∂x)/(∂f/∂y)

Let 

then 



Edwin