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Find partial derivative dz/dx

  1. Nov 4, 2011 #1
    1. The problem statement, all variables and given/known data
    Consider z=sin(x+y+z). This defines z implicitly as a function of x and y. Find an expression for dz/dx



    3. The attempt at a solution
    This was on a test, this is what i did. I got 7/11 pts

    dz/dx = cos(x+y+z)*(1+(dz/dx))

    (dz/dx) / (1 + (dz/dx)) = cos(x+y+z)

    i cant get dz/dx on one side by itself.... help please
     
  2. jcsd
  3. Nov 4, 2011 #2

    I like Serena

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    Hi arl146! :smile:

    Suppose your equation was: x = A * (1 + x)
    Could you solve it then?
     
  4. Nov 4, 2011 #3
    Am I still doing that in terms of a partial derivative? I don't understand your example
     
  5. Nov 4, 2011 #4

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    I've replaced the complex sub expressions in your equation by A and x respectively.
    The result is a simpler equation.
    If you can solve x from that, you should also be able to find dz/dx equivalently.
     
  6. Nov 4, 2011 #5
    Ohh ok. Embarrassing but no I can't haha after I get x/(1+x) idk what to do next
     
  7. Nov 4, 2011 #6

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    The method to solve an equation like this is:
    x = A * (1 + x)
    Remove parentheses: x = A * 1 + A * x
    Subtract A*x on both sides: x - A * x = A * 1
    Extract x, introducing parentheses again: x * (1 - A) = A * 1
    Divide both side by (1-A): x = (A * 1) / (1 - A)

    Does this look familiar?
     
  8. Nov 4, 2011 #7
    Ohh yea ... Don't know why I couldn't do that.. So for my problem, would I get all the z on one side? If I do that I get inverse sin(z)- z=x+y

    Sorry for the lack of normal format, I'm on my phone
     
  9. Nov 4, 2011 #8

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    Hmm, how did you get that?

    It is not right.
    You have a cosine. How did you derive an inverse sine from that?
    And actually, you should not bring all z to one side, but only all (dz/dx).

    Can you perhaps show the steps you took?
     
  10. Nov 4, 2011 #9
    Oh I didn't do the derivative if that yet. But I just did it another way while waiting for you to respond:

    Z=sin(x+y+z)
    z=sin(x)+sin(y)+sin(z)
    z-sin(z)=what's left
    (dz/dx)-cos(z)*(dz/dx)=same stuff
    (dz/dx)(1-cos(z))=sin(x+y)
    dz/dx= (sin(x+y))/(1-cos(z))
     
  11. Nov 4, 2011 #10
    That's prob not right...
     
  12. Nov 4, 2011 #11

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    Sorry, but I'm afraid you can't do:
    Z=sin(x+y+z)
    z=sin(x)+sin(y)+sin(z)

    Your original first step was good.
    That is:
    [itex]z = \sin(x+y+z)[/itex]
    [itex]{\partial z \over \partial x} = \cos(x+y+z) {\partial \over \partial x}(x+y+z)[/itex]
    [itex]{\partial z \over \partial x} = \cos(x+y+z) (1 + {\partial z \over \partial x})[/itex]

    Now you need to solve for [itex]{\partial z \over \partial x}[/itex].
     
  13. Nov 4, 2011 #12
    Oh I got it now. Sorry I'm dumb lol
    dz/dx= cos(x+y+z) / (1+cos(x+y+z))
     
  14. Nov 4, 2011 #13

    Mark44

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    That's definitely not right. sin(x + y + z) ≠ sin(x) + sin(y) + sin(x)
     
  15. Nov 4, 2011 #14

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    Almost! :wink:

    Could you check your plus and minus signs?
     
  16. Nov 4, 2011 #15
    oh yea, i had 1-cos(stuff) not +. again, was on my phone earlier. and thanks!
     
  17. Nov 5, 2011 #16

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    You're welcome! :smile:
     
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