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Partial diff on inverse tan

  1. Jul 28, 2012 #1
    1. The problem statement, all variables and given/known data

    find dz/dy(partial)

    z= tan-1(y/x)



    2. Relevant equations



    3. The attempt at a solution

    z= tan-1(y/x)


    let u=y/x

    z= tan-1(u)

    dz/du = 1/ (1+u2)

    so dz/dy = dz/du (du/dy)

    du/dy = 1/x

    so

    dz/dy = (1/1+u2)(1/x)

    = 1/ 1+ (y2/x2) * 1/x

    = 1/ x + (y2/x2)x)

    = x / x + y2


    but I should be getting...

    dz/dy = x / x2 + y2

    but I found dz/dx the same way and it was the right answer.?

    whoops I made a arithmetic mistake it is correct.

    sorry.
     
    Last edited: Jul 28, 2012
  2. jcsd
  3. Jul 28, 2012 #2

    chiro

    User Avatar
    Science Advisor

    Hey charmedbeauty.

    Going from the appropriate position I get:

    dz/dy = (1/1+u2)(1/x)

    = 1/ (1+ (y2/x2)) * 1/x

    = 1/ (x + (y^2/x^2)x)

    = 1 / (x + y^2/x)

    = 1/ [(x^2 + y^2)/x] ( since x*x/x = x^2/x )

    = x / [x^2 + y^2]
     
  4. Jul 28, 2012 #3
    Your calculation is ok except the last line
     
  5. Jul 28, 2012 #4
    yep for some reason I thought 1/x+(y2/x) = x/ x+y2 but really it did = x/ x2+y2

    Thanks.
     
  6. Jul 28, 2012 #5

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You did not use the distributive law in the denominator: [itex](x+ y^2/x)x= x^2+ y^2[/itex]
     
  7. Jul 28, 2012 #6
    yeah I understand.
     
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