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Derivative of Trig functions problem.

  1. Mar 2, 2013 #1
    1. The problem statement, all variables and given/known data

    If x = asecθ, y =btanθ show that

    dy/dx = (b/a) cosecθ and d2y/dx2 = (-b/a2)cot^3θ





    The attempt at a solution

    I got the 1st part

    dy/dx = (dy/dθ) * (dθ/dx) = bsec^2θ x 1/(secθtanθ)= (b/a) cosecθ

    Now I tried differentiating a 2nd time and I don't get the answer they want me to get. I
    even tried finding d^2y/dθ and dx^2/dθ

    then used the chain rule still didn't get it.

    Please guide me someone :S
     
  2. jcsd
  3. Mar 2, 2013 #2

    Dick

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    How did you use the chain rule? d/dx=(dθ/dx)*(d/dθ). Apply that to the first derivative.
     
  4. Mar 2, 2013 #3
    I used the chain rule to get (b/a) cosecθ

    but I can't get the 2nd derivative.
     
  5. Mar 2, 2013 #4

    Dick

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    Use it again to get the second derivative.
     
  6. Mar 2, 2013 #5
    Oh just do dy/dx times dy/dx?
     
  7. Mar 2, 2013 #6

    Dick

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    No. You do d/dx(dy/dx). Take the expression you got for dy/dx and find d/dx of it again. The same thing you did to get the first derivative from y.
     
  8. Mar 2, 2013 #7
    [tex]\frac{d(\frac{dy}{dx})}{dx}[/tex]

    Maybe that makes it easier to see than Dick wrote it.
     
  9. Mar 2, 2013 #8
    But if I do that I get (d)/(dx)((b csc(x))/a) = -(b cot(x) csc(x))/a
     
  10. Mar 2, 2013 #9

    Dick

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    It's not csc(x). It's csc(θ). Look at it this way. Suppose you were given (b/a) cosecθ (the first derviative) and wanted to find d/dx. That's the same sort of exercise you did to get the first part.
     
  11. Mar 2, 2013 #10
    oh it's d(dydx)/dx = d(dy/dx)/dθ * dθ/dx? This should give me the answer right?
     
  12. Mar 2, 2013 #11

    Dick

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    Yes, that should give you the right answer.
     
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