Question

How do you implicitly differentiate `1=-3xy^2-xy`?

Answer 1

`y'=(-(3y^2+y))/((6xy+x)`

Explanation:

`D(-3xy^2-xy)=D(1)`

`:.-3[x.2yy'+y^2]-[xy'+y]=0`

`:.-6xy.y'-3y^2-xy'-y=0`

`:.-y'(6xy+x)=3y^2+y`

`:.y'=(-(3y^2+y))/((6xy+x))`