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    ((a^2-a+0.5)^0.5)/3?

    I try using the chain rule, but just end up with a mess. Do I ignore the /e when differentiating the top?
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    Is that (a^2 -a + 0.5)^\frac{0.5}{3} ? or \frac{(a^2 -a + 0.5)^(\frac{1}{2})}{3}
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    (Original post by JumpingJonny)
    Is that (a^2 -a + 0.5)^\frac{0.5}{3} ? or \frac{(a^2 -a + 0.5)^0.5}{3}
    The latter. It's everything divided by 3.
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    You can take out the 1/3 as a constant: \frac{d}{dx} \frac{(a^2-a+0.5)^{1/2}}{3} = \frac {1}{3} \frac{d}{dx} (a^2-a+0.5)^{1/2}

    Then use the chain rule with the substitution  u =  a^2-a+0.5
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     y = \dfrac{(a^2-a+0.5)^{0.5}}{3}

    Let u = a^2-a+0.5

     \therefore y = \dfrac{u^{0.5}}{3}

    Note that  \dfrac{\mathrm{d}y}{\mathrm{d}a} = \dfrac{\mathrm{d}y}{\mathrm{d}u} \times \dfrac{\mathrm{d}u}{\mathrm{d}a}

    Well, you can find  \dfrac{\mathrm{d}y}{\mathrm{d}u} and  \dfrac{\mathrm{d}u}{\mathrm{d}a}
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    (Original post by EierVonSatan)
    You can take out the 1/3 as a constant: \frac{d}{dx} \frac{(a^2+a+0.5)^{1/2}}{3} = \frac {1}{3} \frac{d}{dx} (a^2+a+0.5)^{1/2}

    Then use the chain rule with the substitution  u =  a^2+a+0.5
    This, I had to go get dinner so was beaten to it...again :rolleyes:
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    Should I get

    (2a-1)/(6(a^2-1+0.5)^0.5)

    ??
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     \dfrac{(a^2-a+0.5)^0^.^5}{3}

     u = (a^2-a+0.5)}

     \dfrac{du}{dx} = 2a - 1

     \dfrac{dy}{du} = (\dfrac{1}{2})(\dfrac{1}{3})u^-^0^.^5


     \dfrac{du}{dx} * \dfrac{dy}{du}

     (2a - 1)(\dfrac{1}{6})u^-^0^.^5

     \dfrac{(2a - 1)}{6}(a^2 - a + 0.5)^-^0^.^5=

    \dfrac{dy}{dx}= \dfrac{(2a^3 - 3a^2 - 0.5)^-^0^.^5}{6}

    Im not sure if this is right but if it is, can you just leave it in this form?:

     \dfrac{dy}{dx}= \dfrac{(2a - 1)(a^2 - a + 0.5)^-^0^.^5}{6}
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    (Original post by samir12)
     \dfrac{(a^2-a+0.5)^0^.^5}{3}

     u = (a^2-a+0.5)}

     \dfrac{du}{dx} = 2a - 1

     \dfrac{dy}{du} = (\dfrac{1}{2})(\dfrac{1}{3})u^-^0^.^5


     \dfrac{du}{dx} * \dfrac{dy}{du}

     (2a - 1)(\dfrac{1}{6})u^-^0^.^5

     \dfrac{(2a - 1)}{6}(a^2 - a + 0.5)^-^0^.^5=

    \dfrac{dy}{dx}= \dfrac{(2a^3 - 3a^2 - 0.5)^-^0^.^5}{6}
    Correct me if I'm wrong but I don't think you can expand the bracket in the last step...?
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    (Original post by Hippysnake)
    Should I get

    (2a-1)/(6(a^2-a+0.5)^0.5)

    ??
    I assume thats a typo, so yes
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    Well \dfrac{\mbox{d}}{\mbox{d}x} \left[ \dfrac{\left(a^2 - a + \frac{1}{2} \right)^{1/2}}{3} \right] = 0, if that counts for anything. If you're differentiating with respect to a, on the other hand, then the chain rule works. Use y = \dfrac{u^{1/2}}{3},\ u = a^2 - a + \dfrac{1}{2} and it's fairly straightforward.

    (Original post by Hippysnake)
    Correct me if I'm wrong but I don't think you can expand the bracket in the last step...?
    You're right; it should have been \dfrac{2a - 1}{6\sqrt{a^2 - a + \frac{1}{2}}}
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    (Original post by Hippysnake)
    Correct me if I'm wrong but I don't think you can expand the bracket in the last step...?
    You can't.
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    (Original post by EierVonSatan)
    I assume thats a typo, so yes
    It's not a typo...I don't think so anyway



    Eitherway, now I'm stuck on this step-


    -5a^(-6) + (^that equation) = 0


    Looks mighty...
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    I am sorry the typing is so big. I would prefer it were smaller.
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    (Original post by Hippysnake)
    It's not a typo...I don't think so anyway
    So you did mean \frac{2a-1}{6\sqrt{a^2-1+0.5}}?
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    (Original post by EierVonSatan)
    So you did mean \frac{2a-1}{6\sqrt{a^2-1+0.5}}?
    no.....?
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    (Original post by EierVonSatan)
    So you did mean \frac{2a-1}{6\sqrt{a^2-1+0.5}}?
    Why would there be a 1 there? There shouldn't be (and even if there was, -1 + 0.5 = -0.5).
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    anyway guys, could you help with the next part?

    -5a^(-6) + ^^that equation = 0?
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    (Original post by Hippysnake)
    Should I get

    (2a-1)/(6(a^2-1+0.5)^0.5)

    ??
    (Original post by EierVonSatan)
    (Original post by Hippysnake)
    Should I get

    (2a-1)/(6(a^2-a+0.5)^0.5)

    ??
    I assume thats a typo, so yes
    (Original post by Hippysnake)
    It's not a typo...I don't think so anyway
    So it was a typo
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    (Original post by Hippysnake)
    anyway guys, could you help with the next part?

    -5a^(-6) + ^^that equation = 0?
    Which equation? The original one, or the differentiated one? Either way it seems like a fairly ridiculous equation to have to solve. You'd have to move the \dfrac{-5}{a^6} to the other side, square both sides and multiply through by a^{12} to start with... and that would be near impossible to solve without iteration or Wolfram|Alpha.
 
 
 
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