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Math Help - logarithmic differentiation

  1. #1
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    logarithmic differentiation

    if y=x^cosx, find dy/dx.

    help please...Im not sure if I did it right.
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  2. #2
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    Quote Originally Posted by edge View Post
    if y=x^cosx, find dy/dx.

    help please...Im not sure if I did it right.
    "Im not sure if I did it right" means you have done something. Show what you did and we can tell you whether it is right or wrong.
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  3. #3
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    lny=lnx^cosx
    lny=cosxlnx
    y'/y=-sinxlnx+cosx(1/x)
    y'=y(-sinxlnx+cosx/x)
    y'=x^cosx (-sinxlnx+cosx/x)

    is that right...and can I simplify it anymore?
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  4. #4
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    Quote Originally Posted by edge View Post
    lny=lnx^cosx
    lny=cosxlnx
    y'/y=-sinxlnx+cosx(1/x)
    y'=y(-sinxlnx+cosx/x)
    y'=x^cosx (-sinxlnx+cosx/x)

    is that right...and can I simplify it anymore?
    Do you HAVE to use logarithmic differentiation? You can just use chain rule...

    y = x^{\cos{(x)}}.

    Let u = \cos{(x)} so that y = x^u.


    \frac{du}{dx} = -\sin{(x)}

    \frac{dy}{du} = ux^{u - 1} = \cos{(x)}\,x^{\cos{(x)}-1}

    So \frac{dy}{dx} = -\sin{(x)}\cos{(x)}\,x^{\cos{(x)}-1}.
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  5. #5
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    I'm not sure, does what I did work though?
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  6. #6
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    Quote Originally Posted by edge View Post
    I'm not sure, does what I did work though?
    It looks fine.

    Perhaps you'd like to spend some time proving that the two answers are indeed equivalent...
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  7. #7
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    Quote Originally Posted by Prove It View Post
    Do you HAVE to use logarithmic differentiation? You can just use chain rule...

    y = x^{\cos{(x)}}.

    Let u = \cos{(x)} so that y = x^u.


    \frac{du}{dx} = -\sin{(x)}

    \frac{dy}{du} = ux^{u - 1}
    No!!! This is only true for u a constant!

     = \cos{(x)}\,x^{\cos{(x)}-1}

    So \frac{dy}{dx} = -\sin{(x)}\cos{(x)}\,x^{\cos{(x)}-1}.
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  8. #8
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    Quote Originally Posted by Prove It View Post
    It looks fine.

    Perhaps you'd like to spend some time proving that the two answers are indeed equivalent...
    He can't- they are not the same. His answer is correct, yours is wrong.
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