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January 31, 2015

January 31, 2015

Posted by **Oli** on Monday, February 9, 2009 at 5:18am.

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 6:00amNo. 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 6:43amThanks 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 9:40amYes, it's 9x^2 y^2 where I had a coefficient of 6. Good work!

- Calculus -
**Oli**, Monday, February 9, 2009 at 4:32pmAwesome. Think I've done it!

Cheers for the help.

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