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Derivative of Sine Function Raised to An Exponent

  1. Jun 12, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the derivative of (sin x) ^ ((sin(sin x)))


    2. Relevant equations



    3. The attempt at a solution
    I get sin(sin x) * [(sin x) ^ {(sin(sinx)-1)}* cos x] The cos x isn't part of the exponent

    Is this right? Thanks :)
     
  2. jcsd
  3. Jun 12, 2013 #2

    haruspex

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    No, I think it' more complicated than that. First, I would substitute y = sin(x), and use df/dx = df/dy dy/dx. The tricky part is df/dy. For that, try rewriting f(y) as e^(ln f(y)).
     
  4. Jun 12, 2013 #3
    This is what I have:

    u= sin x
    du= cos x
    f(u) = u ^(sin u) du
    f'(u)= u^((sin u)-1)* (sin (u) + ((u)(ln u)(cos u)))
    And then just substitute sin x in for u?
     
  5. Jun 13, 2013 #4

    haruspex

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    Almost. You have computed df/du ok, but you need df/dx. You mention du = cos(x) (properly, du = cos(x)dx), so maybe you hadn't really forgotten that.
     
  6. Jun 13, 2013 #5

    HallsofIvy

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    I would use logarithmic differentiation. If [itex]f(x)= (sin(x))^(sin(sin(x))[itex], then [itex]ln(f(x))= sin(sin(x))ln(sin(x)).

    Now use the product rule and chain rule to find f'(x)/f(x).
     
  7. Jun 13, 2013 #6
    Using the logrithmich method would make the problem easier. Try it, as suggested by one more post
     
  8. Jun 13, 2013 #7
    I'm really confused as to what df/dx would be. If u= sin x, do I rearrange to get the value of x?
     
  9. Jun 14, 2013 #8
    Hint: df/dx = (df/du)*(du/dx). This is called chain rule.

    You better use that lograthmic method, it would be easier. Take
     
  10. Jun 14, 2013 #9

    haruspex

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    In post #3 you can see that Justabeginner has already cracked that part to obtain df/du. All that remains is to use the chain rule, i.e. multiply by du/dx.
     
  11. Jun 14, 2013 #10

    HallsofIvy

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    No, he hasn't. What he had in post #3 was completely wrong.

    In particular, the derivative of [itex]u^{f(u)}[/itex] is NOT [itex]u^{f(u)-1}f'(u)[/itex]

    Instead, taking log of both sides of [itex]y= u^{f(u)}[/itex] gives [itex]ln(y)= f(u)ln(u)[/itex] and differentiating both sides of that, with respect to u, [itex]\frac{1}{y}\frac{dy}{du}= f'(u) ln(u)+ \frac{f(u)}{u}[/itex], [itex]\frac{1}{u^{f(u)}}\frac{dy}{du}[/itex][itex]= f'(u)ln(u)+ \frac{f(u)}{u}[/itex], [itex]\frac{dy}{du}= u^{f(u)}(f'(u)ln(u)+ \frac{f(u)}{u}[/itex]
     
    Last edited: Jun 14, 2013
  12. Jun 14, 2013 #11

    haruspex

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    That isn't what he did.
    which he wrote as
    [itex]\frac{dy}{du}= u^{f(u)-1}(uf'(u)ln(u)+ f(u))[/itex]
     
  13. Jun 15, 2013 #12
    This is the train of thought that I was using as I solved for f'(u).
    And to finish off I'd just multiply f'(u) by du?
     
  14. Jun 15, 2013 #13

    haruspex

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    By du/dx. Maybe that's what you meant.
     
  15. Jun 16, 2013 #14
    Yes sir, that's what I meant. Thank you very much :)
     
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