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Thread: yet another painful derivative

  1. #1
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    yet another painful derivative

    find y'

    y=x^[cos(x)]
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  2. #2
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    Quote Originally Posted by Asuhuman18 View Post
    find y'

    y=x^[cos(x)]
    Use the same approach as explained here: http://www.mathhelpforum.com/math-he...rivatives.html

    If you need more help, please show your work and say where you get stuck.
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  3. #3
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    Quote Originally Posted by Asuhuman18 View Post
    find y'

    y=x^[cos(x)]
    logarithmic differentiation ...

    take the log of both sides

    $\displaystyle \ln{y} = \cos{x} \cdot \ln{x}$

    now take the derivative and solve for y'
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    Quote Originally Posted by mr fantastic View Post
    Use the same approach as explained here: http://www.mathhelpforum.com/math-he...rivatives.html

    If you need more help, please show your work and say where you get stuck.
    so it would be 10e^[ln(x)*cos(x)]
    10e^[ln(x)*cos(x)]* -sin/x
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  5. #5
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    Quote Originally Posted by Asuhuman18 View Post
    so it would be 10e^[ln(x)*cos(x)]
    10e^[ln(x)*cos(x)]* -sin/x
    $\displaystyle u = \cos (x) \cdot \ln (x)$ and you have to use the product rule to differentiate this (the derivative you got for it is wrong).

    And note that in your final answer you should re-write 10e^[ln(x)*cos(x)] in its original form of x^cos(x).
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  6. #6
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    Quote Originally Posted by mr fantastic View Post
    $\displaystyle u = \cos (x) \cdot \ln (x)$ and you have to use the product rule to differentiate this (the derivative you got for it is wrong).

    And note that in your final answer you should re-write 10e^[ln(x)*cos(x)] in its original form of x^cos(x).
    so is the final answer is x^[cos(x)]*-sin(x)*ln(x))+cos(x)*(1/x)?
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  7. #7
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    Quote Originally Posted by Asuhuman18 View Post
    so is the final answer is x^[cos(x)]*(-sin(x)*ln(x))+cos(x)*(1/x))?
    You will need the red brackets.
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