Question: find d(y)/d(x) of y^4+x^3+cos(x+y^2)=0 .
This is an excerise in implicit differentiation.
Implicitly differentating y^4: (y^4)' = (4y^3)*y'
Implicitly differentating cos(x+y^2): (cos(x+y^2))' = -sin(x+y^2)*(x+y^2)' = -sin(x+y^2)*(1+2y*y')
So (4y^3)*y' + 3x^2 - sin(x+y^2)*(1+2y*y') = 0
y'{4y^3 - 2y.sin(x+y^2)} + 3x^2 - sin(x+y^2) = 0
y'{4y^3 - 2y.sin(x+y^2)} = -3x^2 + sin(x+y^2)
y' = {-3x^2 + sin(x+y^2)}/{4y^3 - 2y.sin(x+y^2)}
y' = {-3x^2 + sin(x+y^2)}/{2y(2y^2 - sin(x+y^2))}