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
= -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.
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