Results 1 to 4 of 4

Math Help - Implicit Differentiation

  1. #1
    Senior Member topher0805's Avatar
    Joined
    Jan 2008
    From
    Vancouver
    Posts
    336

    Implicit Differentiation

    Find dy/dx by implicit differentiation.

    tan(x-y) = \frac {y}{(3 + x^2)}

    My Steps:

    \frac {d}{dx} tan(x-y) = sec^2(x-y)(1-y')

    \frac {d}{dx} \frac {y}{(3 + x^2)} = \frac {(3+x^2)y' - y(2x)}{(3+x^2)^2}

    Therefore:

    sec^2(x-y)(1-y') = \frac {(3+x^2)y' - y(2x)}{(3+x^2)^2}

    Now I know that I need to solve for y' but no matter how long I stare at this problem I can't figure it out.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor kalagota's Avatar
    Joined
    Oct 2007
    From
    Taguig City, Philippines
    Posts
    1,026
    Quote Originally Posted by topher0805 View Post
    Find dy/dx by implicit differentiation.

    tan(x-y) = \frac {y}{(3 + x^2)}

    My Steps:

    \frac {d}{dx} tan(x-y) = sec^2(x-y)(1-y')

    \frac {d}{dx} \frac {y}{(3 + x^2)} = \frac {(3+x^2)y' - y(2x)}{(3+x^2)^2}

    Therefore:

    sec^2(x-y)(1-y') = \frac {(3+x^2)y' - y(2x)}{(3+x^2)^2}

    Now I know that I need to solve for y' but no matter how long I stare at this problem I can't figure it out.
    just do the proper grouping and distribution..

    (1-y')\sec^2(x-y) = \sec^2{(x-y)} - y'\sec^2{(x-y)} = \frac {(3+x^2)y' - y(2x)}{(3+x^2)^2}

    (3+x^2)^2\sec^2{(x-y)} - (3+x^2)^2y'\sec^2{(x-y)} = (3+x^2)y' - y(2x).. how about that, can you do the isolation of y'?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member topher0805's Avatar
    Joined
    Jan 2008
    From
    Vancouver
    Posts
    336
    (3+x^2)^2\sec^2{(x-y)} - (3+x^2)^2y'\sec^2{(x-y)} = (3+x^2)y' - y(2x)

    Would then become:

    (3+x^2)^2\sec^2{(x-y)} + y(2x) = (3+x^2)y' + (3+x^2)^2y'\sec^2{(x-y)}

    Which simplifies to:

    y'((3+x^2)+(3+x^2)^2\sec^2{(x-y)}) = (3+x^2)^2\sec^2{(x-y)} + y(2x)

    So the final answer would be:

    y' = \frac {(3+x^2)^2\sec^2{(x-y)} + y(2x)}{(3+x^2)+(3+x^2)^2\sec^2{(x-y)}}

    Is this right?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor kalagota's Avatar
    Joined
    Oct 2007
    From
    Taguig City, Philippines
    Posts
    1,026
    yes..
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: July 26th 2010, 05:24 PM
  2. implicit differentiation
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 1st 2010, 07:42 AM
  3. Implicit Differentiation
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 28th 2010, 03:19 PM
  4. Implicit differentiation
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 26th 2010, 05:41 PM
  5. Implicit Differentiation
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 25th 2008, 07:33 PM

Search Tags


/mathhelpforum @mathhelpforum