1 + x = sin(xy^2)

find dy/dx by implicit differentiation

0 + 1 = cos(xy^2). (x)(2y)dy/dx + (y^2)(1)

1/((x)(2y)dy/dx) = cos(xy^2) + (y^2)

dy/dx = cos (xy^2) + (y^2)....

Can I just divide out the (x)(2y) and leave the dy/dx?

1 answer

you needed brackets like this
0 + 1 = cos(xy^2)*( (x)(2y)dy/dx + (y^2)(1) )
since you need to get at the dy/dx and since it is inside the bracket, you must expand

1 = 2xy(cos(xy^2))dy/dx + y^2(cos(xy^2)
1 - y^2(cos(xy^2) = 2xy(cos(xy^2))dy/dx
dy/dx = (1 - y^2(cos(xy^2))/(2xy(cos(xy^2)))