Question

find dy/dx using implicit differentation for x^3+y^3=3xy^2

Answer 1

differentiate w/respect to x (don't forget the chain rule):\[3x^2+3y^2y'=3y^2+6xyy'\]collect the dy/dx's (or y', same thing):
\[3x^2-3y^2=(6xy-3y^2)y'\]solve for y' and factor out the 3 from numerator and denominator:\[y'=(x^2-y^2)/(2xy-y^2)=dy/dx\]

Answer 2

PS: The product rule is also used on the right side in the first step above.

Answer 3

why didnt the 3y^2 get a y' next to it?

Answer 4

product rule:
(dy/dx)(3xy^2)=3(dx/dx)y^2+3x(d/dx)y^2)

Answer 5

(fg)'=f'g+g'f

Answer 6

still no?
\[D_x(3xy^2)=3(dx/dx)y^2+3x(2y)y'=3y^2+6xyy'\]because dx/dx=1
let me know if you still don't get it.