Question

Use implicit differentiation to find dy/dx given that x^3y-sin2y=4+y^3

Answer 1

(x^3y)'=3dy/dx x^(3y-1)
(-sin2y)'=-2dy/dx cos2y
(4+y^3)'=3dy/dx y^2

factor out dy/dx and you will have your answer

Answer 2

\[x^3y -sin(2y)=4+y^3\]
\[3x^2y+x^3y'-2\cos(y)y'=3y^2y'\]
\[3x^2y=3y^2y'+2\cos(2y)y'-x^3y'\]
\[3x^2y=y'(3y^2+2\cos(2y)-x^3)\]
\[\frac{3x^2y}{3y^2+2\cos(2y)-x^3}=y'\]

Answer 3

check the algebra!

Answer 4

thats what i have satellite thanks :D