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Math Help - Another derivative problem

  1. #1
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    Another derivative problem

    I have problems with finding derivatives using the definition of a derivative when there are square roots involved for some reason. I'm trying to find the derivative of 1/sqrt b and keep getting the same wrong answer, could someone set me straight about how to approach such a problem?
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  2. #2
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    You would look at it like:

    f(b) = b^{-1/2}

    As a square root has an exponent of (1/2), and it being in the denominator makes it negative. From there, it's simple:

    f'(b) = -\frac{1}{2}b^{-3/2}

    or

    f'(b) = -\frac{1}{2\sqrt{b^3}}


    If you mean the limit definition, then:

    f'(b) = \frac{f(b+\Delta b)-f(\Delta b)}{\Delta b}

    as delta-b approaches 0.

    &

    f(b) = b^{-1/2}

    so:

    f'(b) = \frac{(b+\Delta b)^{-1/2}-(\Delta b)^{-1/2}}{\Delta b}


    And then the (-1/2) exponents kind of lose me. Sorry. I feel dumb but I don't remember doing this.
    Last edited by Fire Mage; October 22nd 2008 at 07:02 PM.
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  3. #3
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    thanks that actually cleared it up quite a bit for me.
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  4. #4
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    Alright then!

    I have no clue what to do there really, or at least I'm really tired.
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