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Use implicit differentiation to find dy/dx given that x^3y-sin2y=4+y^3
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(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
\[x^3y -sin(2y)=4+y^3\] \[3x^2y+x^3y'-2\cos(y)y'=3y^2y'\] \[3x^2y=3y^2y'+2\cos(2y)y'-x^3y'\] \[3x^2y=y'(3y^2+2\cos(2y)-x^3)\] \[\frac{3x^2y}{3y^2+2\cos(2y)-x^3}=y'\]
check the algebra!
thats what i have satellite thanks :D
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