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Partial Derivatives

  1. Oct 7, 2004 #1

    Spectre32

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    Partial Derivatives of this(respect to x,y).

    Code (Text):
    Z = (x+y) Sec(xy).
    Would my first move be to multiply the
    Code (Text):
    (x+y)
    tot he other side? If so I'm algerba is a bit sketchy :uhh: , how would it be done.
     
    Spectre32, Oct 7, 2004
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  3. Oct 7, 2004 #2

    faust9

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    You can algebraically expand the quantity (x+y), or you can use the product rule for derivatives. Either method will get you the result. I'd keep it the way it is personally because if you expand the quantity you'll have to use the product rule twice.

    Good luck.
     
    faust9, Oct 7, 2004
  4. Oct 7, 2004 #3

    Spectre32

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    Ok so would I get something like this:

    [(x+y)+((sin(xy)*y)/(cos(xy)^2))] - [(1+y)+sec(xy)] ??
     
    Spectre32, Oct 7, 2004
  5. Oct 7, 2004 #4

    faust9

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    No, how did you get that? Laptop battery almost dead so I'll not respond to this until later tonight BTW. When doing a partial w/ respect to x, y becomes a constant and vice versa.

    Do this: (x+1)sec ((1)(x))d/dx

    What would that look like?
     
    faust9, Oct 7, 2004
  6. Oct 7, 2004 #5

    Spectre32

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    (cos(x) * (x+1)*sin(x))/(Cos(x)^2)
     
    Spectre32, Oct 7, 2004
  7. Oct 7, 2004 #6

    Spectre32

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    Whoops.... Now I see what I did wrong... wow i'm retarded. I forgot to go through and Differentate Y and then figure it out. But I still think I did it wrong. Sec[x] + (1 + x) Sec[x] Tan[x]
     
    Last edited: Oct 7, 2004
    Spectre32, Oct 7, 2004
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