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Math Help - Question on having the root of a variable in denominator

  1. #1
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    Question Question on having the root of a variable in denominator

    Hi, Ive just been learning how to do some derivatives and have become stuck on a question because i think I dont understand some algebra.

    ok question is finding the derivative

    v=1/SQRT(u)
    d/du 1/SQRT(u)?

    the answers say -1/(2u)(SQRTu)

    I know that 1/SQRT(u) = SQRT(u)^-1 but to get u on its own would it then become u^-1.5? or u^-0.5? And I'm wondering is the answer a typo? I thought I knew the power rule and cant see how that answer can e right. I know this is calc now but Im posting here as Im missing some simple algebra/arithmetic
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  2. #2
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    Re: Question on having the root of a variable in denominator

    Quote Originally Posted by camjerlams View Post
    Hi, Ive just been learning how to do some derivatives and have become stuck on a question because i think I dont understand some algebra.

    ok question is finding the derivative

    v=1/SQRT(u)
    d/du 1/SQRT(u)?

    the answers say -1/(2u)(SQRTu)

    I know that 1/SQRT(u) = SQRT(u)^-1 but to get u on its own would it then become u^-1.5? or u^-0.5? And I'm wondering is the answer a typo? I thought I knew the power rule and cant see how that answer can e right. I know this is calc now but Im posting here as Im missing some simple algebra/arithmetic
    All your considerations and calcuations are correct.

    v(u)=\frac1{\sqrt{u}}=u^{-\frac12}

    You differentiated the function correctly:

    v'(u)=-\frac12 \cdot u^{-\frac32} = -\frac1{2 u^{\frac32}}= -\frac1{2 u^{1+\frac12}}=-\frac1{2 u \cdot \sqrt{u}}
    Thanks from camjerlams
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  3. #3
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    Re: Question on having the root of a variable in denominator

    Ah I see, great thankyou.
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