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Math Help - Need refresher on derivatives

  1. #1
    Newbie
    Joined
    Aug 2009
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    Need refresher on derivatives

    I am taking a partial differential class this fall, unfortunately it has been about 12 years since I have had calculus and I have been trying to refresh my memory, but there are a couple of things I just need verification for a few things.

    Given:

    u(x,y,z)=A*sin(ax)*cos(by)*sinh(cz)

    A,a,b,c are constants.

    Find the second derivative with respect to x:

    If I remember correctly, when when taking a derivative with respect to one variable the other variables are treated as constants. In this case when you apply the product rule the first derivative would look like this:

    du/dx=0*sin(ax)*cos(by)*sinh(cz)+A*(a)cos(ax)*cos(by) *sinh(cz)+A*sin(ax)*0*sinh(cz)+A*sin(ax)*cos(by)*0

    du/dx=A*(a)cos(ax)*cos(by)*sinh(cz)

    It would then follow that the second derivative would look like:

    d^2u/dx^2=0*(a)cos(ax)*cos(by)*sinh(cz)+A*(a^2)(-sin(ax))*cos(by)*sinh(cz)+A*(a)cos(ax)*0*sinh(cz)+ A*(a)cos(ax)*cos(by)*0

    d^2u/dx^2=-A*(a^2)sin(ax)cos(by)sinh(cz)

    so basically the second derivative with respect to x looks like:

    u''=-A(a^2)sin(ax)cos(by)sinh(cz)

    So can someone help me out by either confirming my solution or pointing out where I have made mistakes. I appreciate the help.
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  2. #2
    MHF Contributor
    Joined
    Nov 2008
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    France
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    1,458
    Hi

    You are right but you made a useless complication
    u(x,y,z)=A\:\sin(ax)\:\cos(by)\:\sinh(cz)

    As you said you must consider \cos(by)\:\sinh(cz) as constant

    The derivative of kf (with k constant and f function) is kf'

    You can consider u(x,y,z)=B\:\sin(ax) where B=A\:\cos(by)\:\sinh(cz)

    Therefore the partial derivative of u with respect to x is

    \frac{\partial u(x,y,z)}{\partial x}=Ba\:\cos(ax)

    and the second derivative is

    \frac{\partial^2 u(x,y,z)}{\partial x^2}=-Ba^2\:\sin(ax) = -Aa^2\:\sin(ax)\:\cos(by)\:\sinh(cz)

    You do not need to use the product rule
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  3. #3
    Newbie
    Joined
    Aug 2009
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    Thanks, I follow the logic for simplifying the problem, but until I restore my confidence in applying all the rules correctly I wanted to do everything the long way. I appreciate the help and the confirmation, that goes a long way to helping to restore my confidence!
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