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Finding the derivative of the function

  1. Jan 24, 2009 #1
    1. The problem statement, all variables and given/known data

    f(x) = sin(5x) / cos(x^2)



    2. Relevant equations



    3. The attempt at a solution

    I got 25Cosx-Sin 5xCosx / cos x^3

    Would this be the correct answer?

    Thanks!
     
  2. jcsd
  3. Jan 24, 2009 #2

    Dick

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    Nope. Warm up with finding the derivatives of sin(5x) and cos(x^2). What are they? Now apply the quotient rule.
     
  4. Jan 24, 2009 #3
    I have (cos5x)(5)(cosx^2) - (sin5x)(2Cosx) all divided by Cosx^4

    Right track?
     
  5. Jan 24, 2009 #4

    Dick

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    It's getting there. But what's the derivative of cos(x^2)? Chain rule. And cos(x^2)^2 is not cos(x^4).
     
  6. Jan 24, 2009 #5
    Ahh I left out the -sin x

    Okay so I should have

    (cos5x)(5)(Cosx^2) - (sin5x)(2Cosx)(-sinx) / Cosx^4

    ?
     
  7. Jan 24, 2009 #6

    Dick

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    Why don't you just do the pieces first? Derivative of cos(x^2)? You aren't going to get anywhere without that. Or did you actually mean to write cos(x)^2 in the denominator instead of cos(x^2)?
     
  8. Jan 24, 2009 #7
    The derivative of cos x^2 is -2cosxsinx
     
  9. Jan 24, 2009 #8

    Dick

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    Again, is that (cos(x))^2 or cos(x^2)? They are two different things. Correct if it's the first. Use parentheses!
     
  10. Jan 24, 2009 #9
    Its the 2nd one. And I got what your saying. for Cos(x^2) I got -2xsinx^2
     
  11. Jan 24, 2009 #10

    Dick

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    Write that as -2xsin(x^2) and I'll be happier. And cos(x^2)^2 still isn't cos(x^4). Two other different things.
     
  12. Jan 24, 2009 #11
    Does the same thing apply for Sin (5x) that this is different than Sin 5x?
     
  13. Jan 24, 2009 #12
    Is the denominator 2Cos(x^2)?
     
  14. Jan 24, 2009 #13

    Dick

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    When in doubt use parentheses. sin 5x could mean either sin(5)*x or sin(5x). Better to be safe than sorry.
     
  15. Jan 24, 2009 #14

    Dick

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    Geez. The quotient rule says (u/v)'=(u'*v-v'*u)/v^2. v is cos(x^2). v^2 isn't 2*cos(x^2).
     
  16. Jan 24, 2009 #15
    So I got

    (5)Cos (5x)Cos(x^2) - Sin (5x)-2xSin(x^2) / 2Cos (x^2)

    (x^2) squared is x^4, what is Cos (x^2)^2 then?
     
  17. Jan 24, 2009 #16

    Dick

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    (cos(x^2))^2 is different from cos((x^2)^2). One is the square of cos(x^2). The other is cos(x^4). They are different. If you don't believe me pick a number and put it into your calculator. Which do you want?
     
  18. Jan 24, 2009 #17
    After doing some review, I got this as a final answer as I dont think I can simplify it anymore.

    5Cos(5x)Cos(x^2) + 2x Sin(5x)Sin(x^2) / Cos^2(x^2)
     
  19. Jan 24, 2009 #18

    Mark44

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    As Dick asked at least a couple of times, use parentheses. Your last answer would be correctly interpreted as
    [tex]
    5cos(5x)cos(x^2) + \frac{2x sin(5x)Sin(x^2)}{cos^2(x^2)}
    [/tex]
    but I don't think that's what you had in mind.
     
  20. Jan 25, 2009 #19
    I see what your saying. I meant for all of that to be divided by Cos^2(x^2)
     
  21. Jan 25, 2009 #20

    Dick

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    Then I think you've finally got it.
     
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