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Calculus 3: Chain rule

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

    If x=yz and y=2sin(y+z), find dx/dy

    2. Relevant equations

    Chain rule

    3. The attempt at a solution

    From y = 2sin(y+z) we get
    dz/dy= (1-2cos(y+z))/(2cos(y+z))
    dz/dy=((1/2)sec(y+z) - 1)

    dx/dy = ∂x/∂y + ∂x/∂z dz/dy
    = z + y ((1/2)sec(y+z) - 1)
    = z - y - (1/2) sec(y+z)

    But the answer is z-y+tan(y+z) (Mathematical methods in the physical sciences, Boas, Ch. 4 Sec 7 Problem 1)

    What is going wrong?
     
  2. jcsd
  3. Jun 1, 2012 #2

    tiny-tim

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    Hi AlonsoMcLaren! :smile:
    Nooo :redface:
     
  4. Jun 1, 2012 #3
    So how to calculate dz/dy?
     
  5. Jun 1, 2012 #4

    tiny-tim

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    one line at a time, for a start! :rolleyes:

    show us your first line :smile:
     
  6. Jun 1, 2012 #5
    y - 2sin(y+z) = 0

    Let f=y - 2sin(y+z)
    Then ∂f/∂y = 1 - 2cos(y+z)
    ∂f/∂z = -2cos(y+z)

    0 = df = (∂f/∂y)dy+(∂f/∂z)dz
    = (1-2cos(y+z)) dy - 2 cos(y+z) dz

    dz/dy = (1-2cos(y+z))/(2cos(y+z))
     
  7. Jun 1, 2012 #6
    First of all differentiate y=2sin(y+z) with respect to y. From chain rule, we obtain

    $$ 1=2cos(y+z) $$

    Now solve for y and then I think you can solve it.
     
  8. Jun 1, 2012 #7

    tiny-tim

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    oops!

    sorry, that is right :redface:

    you've just left out a y when you expanded this bracket …
     
  9. Jun 1, 2012 #8
    Re: oops!

    So how does it lead to z-y+tan(y+z)?
     
  10. Jun 1, 2012 #9

    tiny-tim

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    x/dy = ∂x/∂y + ∂x/∂z dz/dy

    = z + y ((1/2)sec(y+z) - 1)

    = z - y - (1/2) y sec(y+z)
    = … ? :smile:
     
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