How do you implicitly differentiate #-1=xy^2+2x^2y-e^ysin(3x+y) #?

1 Answer

#y'=(3 e^y*cos (3x+y)-4xy -y^2)/(2xy+2x^2-e^y*sin (3x+y)-e^y*cos(3x+y))#

Explanation:

From the given #-1=x y^2+2x^2 y -e^y*sin (3x+y)#
Differentiate both sides of the equation with respect to #x#

it goes like this

#d/dx(-1)=d/dx(x y^2)+d/dx(2 x^2 y)-d/dx(e^y* sin (3x+y))#

#0=x*2y y'+y^2*1+2[x^2 y'+y*2x]-[e^y*y'*sin(3x+y)+e^y*cos(3x+y)*(3+y')]#

simplification, it follows

#0=2xyy'+y^2+2x^2y'+4xy-e^y sin(3x+y)y'-3 e^y cos(3x+y)-e^y cos(3x+y)y'#

collecting terms with #y'# on one side and transposing the rest of the terms on the other side;

#3e^y cos(3x+y)-4xy-y^2=#
#[2xy+2x^2-e^y sin(3x+y)-e^y cos (3x+y)] y'#

Solve for #y'# by dividing both sides now by #[2xy+2x^2-e^y sin(3x+y)-e^y cos (3x+y)]#

#y'=(3 e^y*cos (3x+y)-4xy -y^2)/(2xy+2x^2-e^y*sin (3x+y)-e^y*cos(3x+y))#