Results 1 to 2 of 2

Math Help - implicit differentiation

  1. #1
    Junior Member
    Joined
    Jun 2008
    From
    Sanger, CA
    Posts
    58

    Exclamation implicit differentiation

    I'm really lost on how to get the derivative of this equation,

    x*sin(x*y)=x^2+1

    I know that the derivative of a sin is cos, but the problem I am having with this equation is getting the derivative of the xy because I don't know what I'm suppose to do with it.

    I know that I'm suppose to use the product rule, but do I also need to use a chain rule or something.... confused...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,829
    Thanks
    1602
    Quote Originally Posted by katchat64 View Post
    I'm really lost on how to get the derivative of this equation,

    x*sin(x*y)=x^2+1

    I know that the derivative of a sin is cos, but the problem I am having with this equation is getting the derivative of the xy because I don't know what I'm suppose to do with it.

    I know that I'm suppose to use the product rule, but do I also need to use a chain rule or something.... confused...
    It will help if you divide both sides by x.

    Then you have

    \sin{(xy)} = x + x^{-1}

    \frac{d}{dx}[\sin{(xy)}] = \frac{d}{dx}(x + x^{-1})

    \frac{d}{dx}(xy)\cdot \frac{d}{du}(\sin{u}) = 1 - x^{-2}, where u = xy

    \left(x\frac{dy}{dx} + y\right)\cos{u} = 1 - \frac{1}{x^2}

    \left(x\frac{dy}{dx} + y\right)\cos{(xy)} = \frac{x^2 - 1}{x^2}

    x\frac{dy}{dx} + y = \frac{x^2 - 1}{x^2\cos{(xy)}}

    x\frac{dy}{dx} = \frac{x^2 - 1}{x^2\cos{(xy)}} - y

    x\frac{dy}{dx} = \frac{x^2 - 1 - x^2y\cos{(xy)}}{\cos{(xy)}}

    \frac{dy}{dx} = \frac{x^2 - 1 - x^2y\cos{(xy)}}{x\cos{(xy)}}.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Implicit Differentiation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 12th 2010, 03:10 PM
  2. Implicit differentiation?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 6th 2010, 09:51 AM
  3. Replies: 2
    Last Post: July 26th 2010, 06:24 PM
  4. implicit differentiation help
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 27th 2010, 10:46 PM
  5. implicit differentiation
    Posted in the Calculus Forum
    Replies: 3
    Last Post: December 3rd 2007, 07:12 PM

Search Tags


/mathhelpforum @mathhelpforum