Posted by Oli on Monday, February 9, 2009 at 5:18am.
Hi,
I'm having trouble using implicit differentiation to determine dy/dx in the form dy/dx = f(x,y) for
sin(2x+3y)=3x^3y^2+4
Do I make it sin(2x+3y)3x^3y^2=4 then differentiate to get
2cos(2x+3y)9x^2*2y=0 ?
I'm a little lost...
Any help appreciated.

Calculus  drwls, Monday, February 9, 2009 at 6:00am
No. Your first step is IK but unecessary, and your second differentiation step is wrong.
Differentiate both sides of the equation with repect to x, remembering that y is a function of x. Use the "chain rule" when necessary.
cos(2x+3y)*(2 + 3dy/dx) = 6y^2x^2 +6x^3*y*dy/dx
Rearrange and solve for dy/dx
dy/dx[3cos(2x+3y)6x^3 y]
= 2cos(2x+3y) 6y^2x^2
One more step 
Calculus  Oli, Monday, February 9, 2009 at 6:43am
Thanks for your reply.
I tried again and got
cos(2x+3y)*(2 + 3dy/dx = 9x^2y^2 + (6x^3*y dy/dx)
Looking at your answer it seems wrong also.
Should it not be 9y^2x^2... on the 1st part of the right hand side of your equation? Or am I missing something here?
Getting stuck on the rearranging but will it another try 
Calculus  drwls, Monday, February 9, 2009 at 9:40am
Yes, it's 9x^2 y^2 where I had a coefficient of 6. Good work!

Calculus  Oli, Monday, February 9, 2009 at 4:32pm
Awesome. Think I've done it!
Cheers for the help.