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Thread: Differentiation from First Principals

  1. September 6th 2010, 08:08 PM #1
    liveboon
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    Cool Differentiation from First Principals

    ok i dont know how to write square root so '\' means 'square root'

    Q1. Differentiate by using the Chain Rule

    f(x)=\2x-1

    Square root of 2x-1

    Q2. Differentiate by First Principals

    f(x)=\2x-1

    I already know that the answers should be the same but i cant work it out using first principals and keep ending up with f'(x)=1
    Last edited by mr fantastic; September 6th 2010 at 08:19 PM. Reason: Re-titled.
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  2. September 7th 2010, 01:11 AM #2
    pickslides
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    What are your answers?

    Spoiler:

    \displaystyle \frac{\sqrt{2(x+h)-1)}-\sqrt{2x-1}}{h}

    \displaystyle \frac{1}{h}(\sqrt{2(x+h)-1)}-\sqrt{2x-1})

    \displaystyle \frac{1}{h}(\sqrt{2(x+h)-1)}-\sqrt{2x-1})\times \frac{\sqrt{2(x+h)-1)}+\sqrt{2x-1}}{\sqrt{2(x+h)-1)}+\sqrt{2x-1}}

    \displaystyle \frac{1}{h}(\sqrt{2(x+h)-1)}-\sqrt{2x-1})\times \frac{\sqrt{2(x+h)-1)}+\sqrt{2x-1}}{\sqrt{2(x+h)-1)}+\sqrt{2x-1}}

    \displaystyle \frac{1}{h}\left(\frac{2x+2h-1-2x+1}{\sqrt{2(x+h)-1)}+\sqrt{2x-1}}\right)

    \displaystyle \frac{1}{h}\left(\frac{2h}{\sqrt{2(x+h)-1)}+\sqrt{2x-1}}\right)

    \displaystyle \frac{2}{\sqrt{2(x+h)-1)}+\sqrt{2x-1}}

    Now

    \displaystyle h\to 0

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