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Math Help - Hard Derivative.

  1. #1
    Member mybrohshi5's Avatar
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    Hard Derivative.

    y=sqrt(10^(5-x))

    find y'

    how would you go about doing this one?

    im studying for my "mastery test" on monday and i have never seen anything raise to a number - x

    could someone show me how to do this please?

    thank you
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by mybrohshi5 View Post
    y=sqrt(10^(5-x))

    find y'

    how would you go about doing this one?

    im studying for my "mastery test" on monday and i have never seen anything raise to a number - x

    could someone show me how to do this please?

    thank you
    Note that \sqrt{10^{5-x}}=10^{\frac{1}{2}(5-x)}. Then use the fact that \frac{\,d}{\,dx}a^u=a^u\ln a\cdot\frac{\,du}{\,dx}.

    Can you take it from here?
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  3. #3
    Member mybrohshi5's Avatar
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    yeah i can. but i seem to get a different answer then the answer of my review sheet.

    i get -10^(.5(5-x)) * ln(10)

    but....

    the answer on the sheet is -10^(.5(5-x)) / 2
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  4. #4
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    Use logarithmic differentiation.
    sqrt of 10^(5-x) is the same as 10^[(5-x)/2]
    Take the natural log of both sides.
    ln y = ln 10^[(5-x)/2]
    ln y = [(5-x)/2] ln 10 because lnx^a = a lnx
    then differentiate:
    (1/y)(dy/dx) = (-1/2)ln10
    solve for dy/dx:
    dy/dx = (-1/2)(ln10)(10^[(5-x)/2])

    edit: The answer sheet is wrong? :P
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  5. #5
    Member mybrohshi5's Avatar
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    haha ok thank you very much =)
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