Question 1208278
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Find dy/dx for x^3+2xy^2 = sin y
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<pre>
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) = {{{(dy)/(dx)}}} 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' = {{{(3x^2 + 2y^2)/(cos(y)-4xy)}}}.    <U>ANSWER</U>


It is the formula which you want to get.
</pre>

Solved.



Having this idea, this technique and this TEMPLATE in your mind, 
you can solve million other similar problems.