Wednesday
March 22, 2017

Post a New Question

Posted by 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 - , 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 - , 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 - , Monday, February 9, 2009 at 9:40am

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

  • Calculus - , Monday, February 9, 2009 at 4:32pm

    Awesome. Think I've done it!

    Cheers for the help.

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question