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Math Help - Partial derivative

  1. #1
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    Partial derivative

    Having a bit of a problem with this partial derivative, i know how to do the problem if the x and y wasn't infront as you could apply the product rule and apply the partial derivative, however having problems in what to do with it, here is the problem:

    f(x,y)=xcosx*coshy+ysinx*sinhy

    Determine partial df/dx, df/dy, second partial df/dx, second partial df/dy and partial d^2f/dxdy

    Many thanks in advance.
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  2. #2
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    This problem may have arisen from a course in DE's, but it's really a calculus question. I'll do the first-order derivatives - see if you can finish with all the second-order derivatives. We have

    f(x,y)=x\cos(x)\cosh(y)+y\sin(x)\sinh(y). Then,

    \dfrac{\partial f}{\partial x}=(1\cdot \cos(x)-x\sin(x))\cosh(y)+y\cos(x)\sinh(y), and

    \dfrac{\partial f}{\partial y}=x\cos(x)\sinh(y)+\sin(x)(1\cdot\sinh(y)+y\cosh(  y)).

    So you can see that I just used the usual product rule from Calc I to differentiate the products.
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  3. #3
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    Quote Originally Posted by Ackbeet View Post
    This problem may have arisen from a course in DE's, but it's really a calculus question. I'll do the first-order derivatives - see if you can finish with all the second-order derivatives. We have

    f(x,y)=x\cos(x)\cosh(y)+y\sin(x)\sinh(y). Then,

    \dfrac{\partial f}{\partial x}=(1\cdot \cos(x)-x\sin(x))\cosh(y)+y\cos(x)\sinh(y), and

    \dfrac{\partial f}{\partial y}=x\cos(x)\sinh(y)+\sin(x)(1\cdot\sinh(y)+y\cosh(  y)).

    So you can see that I just used the usual product rule from Calc I to differentiate the products.
    Fantastic i can see exactly where i went wrong, thanks very much. Just going out and will work on the second derivative later, thanks once more.
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  4. #4
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    You're welcome. Let me know if you have any more difficulties.
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  5. #5
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    Quote Originally Posted by Ackbeet View Post
    You're welcome. Let me know if you have any more difficulties.
    Applied it to other questions and getting them spot on now, however there is a second part where it says:
    Without further explicit differentiation, determine partial (d^4f/dx^4)-(d^4f/dy^4) ive calculated the second partial derivative etc just wondering if there is any trick to it as it states without any further explicit differentiation.
    Any help will be most appreciated.
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  6. #6
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    Is this the same f as in the OP?
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  7. #7
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    If you've calculated f_{xx} and f_{yy} you'll notice that

    f_{xx} + f_{yy} = 0.

    So how does that help with

    f_{xxxx} - f_{yyyy} = 0?

    Spoiler:

    f_{xxxx} - f_{yyyy} =  0 \;\text{can be written as}\;  \left(f_{xx} + f_{yy}\right)_{xx} - \left(f_{xx} + f_{yy}\right)_{yy} = 0 so your f solves the 4th order PDE
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  8. #8
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    Quote Originally Posted by Danny View Post
    If you've calculated f_{xx} and f_{yy} you'll notice that

    f_{xx} + f_{yy} = 0.

    So how does that help with

    f_{xxxx} - f_{yyyy} = 0?

    Spoiler:

    f_{xxxx} - f_{yyyy} =  0 \;\text{can be written as}\;  \left(f_{xx} + f_{yy}\right)_{xx} - \left(f_{xx} + f_{yy}\right)_{yy} = 0 so your f solves the 4th order PDE
    I worked out partal d^2f/dx^2 and partial d^2f/dy^2 added together to be 0. Then wasn't sure where to go from there.
    Many thanks in advance.
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