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Math Help - Differentiate

  1. #1
    Member SengNee's Avatar
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    Differentiate

    Differentiate:
    a) \ln\left({\frac{2}{\sqrt{x}}}\right)
    b) \sin x-2\cos x , x=\frac{\pi}{4}
    Last edited by SengNee; March 23rd 2008 at 06:54 PM.
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    Expand the first one before takin' its derivative. As for the second one it's just a matter of basic derivatives, check it out in your notes.
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  3. #3
    Member SengNee's Avatar
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    Differentiate:
    a) \ln\left({\frac{2}{\sqrt{x}}}\right)
    b) \sin x-2\cos x , x=\frac{\pi}{4}

    a)

    \frac{d}{dx}\left[\ln\left({\frac{2}{\sqrt{x}}}\right)\right]
    =\frac{-\frac{1}{2}(2x^{-\frac{3}{2}})}{2x^{-\frac{1}{2}}}
    =-\frac{1}{2x}

    But the book's answer is \frac{1}{2x} .
    Is my answer wrong?


    b)

    \frac{d}{dx}\left(\sin x-2\cos x\right)
    =\cos x +2\sin x
    =\frac{1}{\sqrt 2}+2\left(\frac{1}{\sqrt 2}\right)
    =\frac{3}{\sqrt 2}

    But the book's answer is -\frac{1}{\sqrt 2} .
    Is my answer wrong again?
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by SengNee View Post
    a)

    \frac{d}{dx}\left[\ln\left({\frac{2}{\sqrt{x}}}\right)\right]
    =\frac{-\frac{1}{2}(2x^{-\frac{3}{2}})}{2x^{-\frac{1}{2}}}
    =-\frac{1}{2x}

    But the book's answer is \frac{1}{2x} .
    Is my answer wrong?


    b)

    \frac{d}{dx}\left(\sin x-2\cos x\right)
    =\cos x +2\sin x
    =\frac{1}{\sqrt 2}+2\left(\frac{1}{\sqrt 2}\right)
    =\frac{3}{\sqrt 2}

    But the book's answer is -\frac{1}{\sqrt 2} .
    Is my answer wrong again?
    I agree with you for the answer to both problems.

    -Dan
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  5. #5
    Flow Master
    mr fantastic's Avatar
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    Quote Originally Posted by topsquark View Post
    I agree with you for the answer to both problems.

    -Dan
    Seconded.
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