Results 1 to 5 of 5

Math Help - Show how this derivative is computed

  1. #1
    Newbie
    Joined
    May 2011
    Posts
    8

    Show how this derivative is computed

    Good morning all, I am following a book which covers aspects of astronomy/cosmology, I'd like to know how the part in the middle is worked out. I have the start and end of the derivation but its quite scant on how it proceeds from point to point, the book simply says using the chain rule "this" becomes "this", basically.

    ∂ρ/∂z

    Where the solution for \rho is:

    \rho (z) = \rho exp (-z^2/2H^2)

    The result is:

    (-z/H^2) \rho

    This is simply for myself and trying to understand how the calculations are derived.

    Thanks in advance

    Rod.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,224
    Thanks
    1791
    Looks to me like you have copied incorrectly- you should not have that \rho on the right side of the equation defining the function \rho(z).

    If \rho(z)= e^{-z^2/(2H^2)}= exp(-z^2/(2H^2)) then we use the chain rule:
    Letting u= -z^2/(2H^2), \rho= e^{u}.

    Then \frac{d\rho}{du}= e^{u}= \rho and \frac{du}{dz}= -2z/(2H^2)= -z/H^2

    By the chain rule then
    \frac{d\rho}{dz}= \frac{d\rho}{du}\frac{du}{dz}= (-z/H^2)\rho(z)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2011
    Posts
    8
    Quote Originally Posted by HallsofIvy View Post
    Looks to me like you have copied incorrectly- you should not have that \rho on the right side of the equation defining the function \rho(z).
    It could probably be my copying, forgive an old man I am not well adjusted to using the symbol format on the forum. When I retired so it seems did my eyesight and finger control.

    I'll try and be more accurate:

    ∂ρ/∂z

    Where the solution for \rho (z) is:

    \rho (z) = \rho{}_{c}   exp(-z^2/2H^2)

    Then it jumps to:

    ∂ρ/∂z = (-2z/2H^2)\rho{}_{c}   exp(-z^2/2H^2) = (-z/H^2) \rho
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    May 2011
    Posts
    8
    OK, after a few more hours and using what HallsofIvy kindly replied with as a start I am still not sure how to derive this:

    \rho (z) = \rho{}_{c}   exp(-z^2/2H^2)

    Then it jumps to:

    ∂ρ/∂z = (-2z/2H^2)\rho{}_{c}   exp(-z^2/2H^2) = (-z/H^2) \rho
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,224
    Thanks
    1791
    Quote Originally Posted by RodBrown View Post
    OK, after a few more hours and using what HallsofIvy kindly replied with as a start I am still not sure how to derive this:

    \rho (z) = \rho{}_{c}   exp(-z^2/2H^2)

    Then it jumps to:

    ∂ρ/∂z = (-2z/2H^2)\rho{}_{c}   exp(-z^2/2H^2) = (-z/H^2) \rho
    Writing u= -z^2/H^2 we have [tex]\rho(u)= \rho_ce^{u}[/itex] so the \frac{d\rho}{du}= \rho_ce^u
    Do you understand that? It is simply asserting that the derivative of the exponential is the exponential and is a defining property of the exponential.

    Also \frac{du}{dz}= -2z/H^2. That is just using the fact that the derivative of z squared is 2z.

    Then, by the chain rule
    \frac{d\rho}{dz}= \frac{d\rho}{du}\frac{du}{dz}= \left(\rho_ze^{u}\right)\left(-2z/H^2\right)
    but that first parentheses is \rho(z) so that
    \frac{d\rho}{dz}= (-2z/H^2)\rho(z)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 12
    Last Post: May 26th 2011, 09:48 AM
  2. Show how this derivative is computed
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 12th 2011, 03:22 AM
  3. Replies: 3
    Last Post: April 27th 2010, 10:55 AM
  4. Show definition of derivative
    Posted in the Calculus Forum
    Replies: 4
    Last Post: January 30th 2010, 12:22 PM
  5. show how the derivative reduces...
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 26th 2009, 07:53 AM

Search Tags


/mathhelpforum @mathhelpforum