Calculus
posted by Oli on .
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.

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 
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 
Yes, it's 9x^2 y^2 where I had a coefficient of 6. Good work!

Awesome. Think I've done it!
Cheers for the help.