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Math Help - derivative of sine function with exponent x

  1. #1
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    derivative of sine function with exponent x

    Okay, so I've tried doing this problem a million times and I can't seem to get the right answer. It seems fairly straight forward.
    Here is the question:
    If f(x)=6(sin(x))x , find f(3)

    So this is what I've tried:
    lny=6xlnsinx
    (1/y)(y')=f'g+fg'
    (1/y)(y')=(6)(lnsinx)+[(6x)((cosx)/(sinx))]
    y'=[(6lnsin(x))+(6xcot(x))]*(6(sin(x))^x)
    So I tried pluggin in 3 to the answer, but it does not give me the correct answer.
    Am I doing something wrong?

    (I've also tried with the assumption that lny=xln6sinx, but that didn't work either.
    Thanx!
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  2. #2
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    There's no reason to take logs, it's just a product of functions, apply the product rule... or is it just that I'm misunderstanding the problem and it's 6(\sin x)^x ?
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  3. #3
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    Ya sorry, the question is actually f(x)=6(sin(x))^x like you said. Thought I typed it like that.. sorry...
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  4. #4
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    Well in that case, it does make sense to apply logs, but make it easier, put (\sin x)^x=e^{x\ln(\sin x)}, now, contemplate its derivative.
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  5. #5
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    uhhhh... sorry, I think you lost me. Where did you get that expression from?
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  6. #6
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    Because of x=e^{\ln x} for x>0.
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  7. #7
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    okay, so I think I found the derivative.
    I got:

    [(lnsinx)+(xcotx)][6e^(xlnsinx)]
    But it still isn't right.
    Did I do this wrong?
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  8. #8
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    Well, having f(x)=6(\sin x)^x=6e^{x\ln(\sin x)}, hence f'(x)=6e^{x\ln(\sin x)}\cdot(x\ln(\sin x))', because of (x\ln(\sin x))'=\ln(\sin x)+x\cot x the derivative equals f'(x)=6(\sin x)^x\cdot(\ln(\sin x)+x\cot x) and we're done.
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  9. #9
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    Thanks. That makes complete sense. I actually did get it right the last time, I was looking at the wrong answer in my textbook. Haha. Thanks for all your help!!
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