Advertisement
Help Keep Boards Alive. Support us by going ad free today. See here: https://subscriptions.boards.ie/.
If we do not hit our goal we will be forced to close the site.

Current status: https://keepboardsalive.com/

Annual subs are best for most impact. If you are still undecided on going Ad Free - you can also donate using the Paypal Donate option. All contribution helps. Thank you.
https://www.boards.ie/group/1878-subscribers-forum

Private Group for paid up members of Boards.ie. Join the club.

Differentiation Question

  • 14-05-2007 06:23PM
    #1
    Sinfonia
    Registered Users, Registered Users 2 Posts: 3,854 ✭✭✭


    Hello,


    I'm puzzled over finding the Derivate for:

    ln(1+e)+w-0.5(1+e)^2


    My lecture notes say:

    b/(1+e) - (1+e) = 0

    (Does this mean we are differentiating with respect to (1+e), or just e?)


    For the last term (after the minus) I understand how that was achieved, and the w disappears, but I don't understand the first part, i.e.

    ln(1+e)

    I know that the derivative for ln(1+e) would be 1/(1+e), but how does the b end up on top, assuming it has to be part of the equation and not just multiplied at the end:confused:

    Thanks in advance


Comments

  • #2
    Irish Wolf
    Registered Users, Registered Users 2 Posts: 2,178 ✭✭✭Irish Wolf


    ln(1+e) = b(ln(1+e))

    => derivative = b(1/(1+e)) = b/(1+e)


  • #3
    Sinfonia
    Registered Users, Registered Users 2 Posts: 3,854 ✭✭✭Sinfonia


    Irish Wolf wrote:
    ln(1+e) = b(ln(1+e))

    => derivative = b(1/(1+e)) = b/(1+e)

    Thanks IW,

    (Just for clarification):
    So the b is just multiplied at the end, and that's all there is to it?


  • #4
    Irish Wolf
    Registered Users, Registered Users 2 Posts: 2,178 ✭✭✭Irish Wolf


    Yeah - it's a constant treated the same as the 0.5 in the final term.


  • #5
    Sinfonia
    Registered Users, Registered Users 2 Posts: 3,854 ✭✭✭Sinfonia


    Brilliant, cheers :)


Advertisement