Results 1 to 3 of 3

Math Help - Implicit Differentiation

  1. #1
    Member
    Joined
    Apr 2009
    Posts
    83

    Implicit Differentiation

    Assuming that the equations define x and y implicitly as differentiable functions x = f(t), y = g(t), find the slope of the curve x = f(t),
    y = g(t) at the given value of t.




    i tried to differential both functions but there are still y and x-variables. can someone guide me through this?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor undefined's Avatar
    Joined
    Mar 2010
    From
    Chicago
    Posts
    2,340
    Awards
    1
    Quote Originally Posted by krzyrice View Post
    Assuming that the equations define x and y implicitly as differentiable functions x = f(t), y = g(t), find the slope of the curve x = f(t),
    y = g(t) at the given value of t.




    i tried to differential both functions but there are still y and x-variables. can someone guide me through this?
    Using the first equation, plug in t = 0 and you will get x = 0.

    Using the second equation, plug in t = 0 and you will get y = 9.

    So when you differentiate implicitly for the first equation, you will have some x's, t's, and dx/dt's. Just solve for dx/dt and don't plug in values yet.

    Then for the second equation you will have y's, t's, and dy/dt's. Solve for dy/dt without plugging in values.

    Then take dy/dt divided by dx/dt to get dy/dx.

    Then plug in x = 0, y = 9, and t = 0 and you should get the proper answer.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member AllanCuz's Avatar
    Joined
    Apr 2010
    From
    Canada
    Posts
    384
    Thanks
    4
    Quote Originally Posted by krzyrice View Post
    Assuming that the equations define x and y implicitly as differentiable functions x = f(t), y = g(t), find the slope of the curve x = f(t),
    y = g(t) at the given value of t.




    i tried to differential both functions but there are still y and x-variables. can someone guide me through this?
    This is okay, consider the following

     y=f(x)= e^{x+5}

    And

     lny = lne^{x+5} = x+5

    If we want to differentiate the above implicitly we get,

     \frac{1}{y} y^{ \prime } = 1

    Which follows

     y^{ \prime} = y

    Notice how  y^{ \prime} = y = f(x)

    This is okay!

      y^{ \prime} = y = e^{x+5}

    We are allowed to have our derivative as a function of our original function, there is no problem.

    Knowing this can you now compute your question?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: July 26th 2010, 06:24 PM
  2. implicit differentiation
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 10th 2010, 06:58 PM
  3. Implicit differentiation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 20th 2009, 07:02 PM
  4. implicit differentiation
    Posted in the Calculus Forum
    Replies: 4
    Last Post: September 12th 2009, 04:15 PM
  5. Implicit Differentiation
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 29th 2007, 06:11 PM

Search Tags


/mathhelpforum @mathhelpforum