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Math Help - yet another painful derivative

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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

    \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
    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
    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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