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Math Help - How'd they differentiate this?

  1. #1
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    Question How'd they differentiate this?

    dR/dh = 0,
    where R is 2 (hH-h2 )1/2
    d [2 (hH-h2 )1/2 ] /dh = 0
    2 x 1/2 (hH-h2)-1/2 x (H-2h) = 0
    (H-2h)/rt (hH-h) = 0
    H-2h=0
    h = H/2


    this is a part of a physics question and any help is appreciated.
    I get it till the 2 x1/2 part, but how did they get the (h-2h) part?
    If someone could explain this to me, I'd be super-grateful!
    Last edited by Booksrock; December 30th 2012 at 09:01 PM. Reason: mistake in the question!
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  2. #2
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    Re: How'd they differentiate this?

    R=2(hH-h^2)^1/2
    Let hH-h^2=t
    R=2t^1/2
    dR/dt=2x(1/2)xt^-(1/2)
    =t^-(1/2)
    t=hH-h^2
    dt/dh=H-2h
    dR/dh=(dR/dt)x(dt/dh)= t^-(1/2)x(H-2h)
    =(H-2h)/(hH-h^2)^1/2
    If we put dR/dh=0
    dR/dh=(H-2h)/(hH-h^2)^1/2=0
    First thing is that denominator should not be zero hence hH-h^2 is non zero
    Therefore h can't be zero and can't be equal to H
    numerator should be zero, thus H=2h
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  3. #3
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    Re: How'd they differentiate this?

    Hello, Booksrock!

    The two h's are confusing.
    I assume that H is a constant . . . Let's call it A.


    \text{Given: }\,R \:=\:2(Ah - h^2)^{\frac{1}{2}}. \quad \text{Solve: }\,\frac{dR}{dh} \,=\,0

    Chain Rule: . \frac{dR}{dh} \;=\;2\cdot\overbrace{\tfrac{1}{2}(Ah - h^2)^{-\frac{1}{2}}}^{\frac{1}{2}\text{ power}}\cdot\underbrace{(A - 2h)}_{\text{"inside"}} \;=\;0
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