Results 1 to 8 of 8

Math Help - Difficult differentiation question

  1. #1
    Senior Member
    Joined
    Feb 2008
    Posts
    297

    Difficult differentiation question

    Let\ x=e^t.\ Show\ that:

    {\frac{d^2y}{dx^2}}={\frac{1}{x^2}} \left(\frac{d^2y}{dt^2}-\frac{dy}{dt}\right)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,110
    Thanks
    986
    Quote Originally Posted by acevipa View Post
    Let\ x=e^t.\ Show\ that:

    d^2y/dx^2={\frac{1}{x^2}}(d^2y/dt^2-dy/dt)
    how is y related to x or to t ?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,829
    Thanks
    1602
    Quote Originally Posted by acevipa View Post
    Let\ x=e^t.\ Show\ that:

    {\frac{d^2y}{dx^2}}={\frac{1}{x^2}}(\frac{d^2y}{dt  ^2}-\frac{dy}{dt}
    I don't think you've given enough information... What is y?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Feb 2008
    Posts
    297
    Quote Originally Posted by Prove It View Post
    I don't think you've given enough information... What is y?
    That's what confused me a little. That's what the question says, unless it's a typo.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Feb 2008
    Posts
    297
    It's just a "show" question. So could it have something to do with starting from \frac{d^2y}{dx^2} and reducing it into a form which only has x and t. Not really sure, how you would do that though.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Apr 2010
    From
    london
    Posts
    11
    this question is a pretty straight forward use of the chain rule in higher derivatives.

    There is no need for you to know the relation between y and t because as you can see in the question this relation is expressed as a differential.

    what you need for solving this is to notice that if x=e^t then t=ln(x)
    then dt/dx= x^(-1) and the second derivative is d^2t/dx^2= -x^(-2)

    now using the chain rule
    d^2y/dt^2= (d^2y/dt^2) * (dt/dx)^2 + (dy/dt) * (d^2t/dx^2)
    its enough that you substitute for the values of dt/dx and d^2t/dx^2 what you derived in the beginning and you're done.


    All it takes is understanding that the chain rule can be used for higher derivatives as well.



    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member
    Joined
    Feb 2008
    Posts
    297
    Quote Originally Posted by schmeling View Post
    this question is a pretty straight forward use of the chain rule in higher derivatives.

    There is no need for you to know the relation between y and t because as you can see in the question this relation is expressed as a differential.

    what you need for solving this is to notice that if x=e^t then t=ln(x)
    then dt/dx= x^(-1) and the second derivative is d^2t/dx^2= -x^(-2)

    now using the chain rule
    d^2y/dt^2= (d^2y/dt^2) * (dt/dx)^2 + (dy/dt) * (d^2t/dx^2)
    its enough that you substitute for the values of dt/dx and d^2t/dx^2 what you derived in the beginning and you're done.


    All it takes is understanding that the chain rule can be used for higher derivatives as well.


    How did you get this line of working?

    d^2y/dt^2= (d^2y/dt^2) * (dt/dx)^2 + (dy/dt) * (d^2t/dx^2)
    Last edited by acevipa; April 24th 2010 at 06:31 AM.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Apr 2010
    From
    london
    Posts
    11
    this is the chain rule for a second derivative of a function. You either need to have a formula booklet for that or use the Faa di Bruno's formula to derive it yourself.
    Its beyond my latex using ability to post the actual derivation so far (ive just joined today) so I assumed that they expect you to know the chain rule for the second derivative rather than derive it but if you want to find out more about how to derive that yourself just read up on Faa di Bruno.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. difficult question!!!!
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: July 17th 2009, 12:29 PM
  2. Replies: 5
    Last Post: November 19th 2008, 10:01 PM
  3. Somewhat difficult question.
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: October 14th 2008, 06:52 AM
  4. Difficult Differentiation
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 21st 2008, 05:37 PM
  5. Difficult differentiation of e questions.
    Posted in the Calculus Forum
    Replies: 3
    Last Post: July 20th 2008, 02:18 AM

Search Tags


/mathhelpforum @mathhelpforum