Question 1208278
Find dy/dx for x^3+2xy^2 = sin y
<pre>
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) 

{{{dy/dx}}}{{{""=""}}} -(∂f/∂x)/(∂f/∂y)

Let {{{"f(x,y)"}}}{{{""=""}}}{{{x^3+2xy^2-sin(y)}}}{{{""=""}}}{{{0}}}

then {{{dy/dx}}}{{{""=""}}}{{{-expr((3x^2+2y^2)/(4xy-cos(y)^""))}}}

{{{dy/dx}}}{{{""=""}}}{{{expr((3x^2+2y^2)/(cos(y)^""-4xy))}}}

Edwin</pre>