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Product rule proof

  1. Sep 22, 2015 #1
    Why wouldn't this work? Screenshot_2015-09-22-16-03-31.png
     
  2. jcsd
  3. Sep 22, 2015 #2

    Orodruin

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    Because
    $$
    \lim_{\Delta x\to 0}\left(\frac{\Delta(g(x)h(x))}{\Delta x}\right) \neq \lim_{\Delta x\to 0}\left(\frac{\Delta g(x)}{\Delta x}\right)\lim_{\Delta x\to 0}\left(\frac{\Delta h(x)}{\Delta x}\right).
    $$
    It is unclear why you think these expressions would be equal on quite a few levels.
     
  4. Sep 22, 2015 #3
    Isn't that a limit law though? Screenshot_2015-09-22-16-11-32.png
     
  5. Sep 22, 2015 #4

    Orodruin

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    You are doing something completely different from that law. Not only are you not computing ##\Delta(g(x) h(x))## correctly, you are also introducing an arbitrary extra ##\Delta x## in the denominator.

    Also, you should stop attaching images like these. They are going to be impossible to read on most mobile devices and impossible to quote properly. There is a tutorial on how to write proper forum maths here: https://www.physicsforums.com/help/latexhelp/
     
  6. Sep 22, 2015 #5
    I understand that I put in an extra Δx. But how is Δ(g(x)h(x)) beung computed incorrectly?
     
  7. Sep 22, 2015 #6

    Orodruin

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    You are assuming ##\Delta(g(x)h(x)) = (\Delta g(x))(\Delta h(x))## to leading order in ##\Delta x##:
     
  8. Sep 22, 2015 #7
    Oh ok..thanks!
     
  9. Sep 22, 2015 #8

    Mark44

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    I agree completely. The image you posted in #3 can be done right here in the text input pane.
     
  10. Sep 25, 2015 #9

    Svein

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    The standard proof: Let [itex] f(x)=u(x)\cdot v(x)[/itex]. Then [itex]f(x+h)=u(x+h)\cdot v(x+h) [/itex]. This means that [itex]f(x+h)-f(x)=u(x+h)\cdot v(x+h) - u(x)\cdot v(x) [/itex]. Now add and subtract [itex]u(x+h)v(x) [/itex]: [itex] f(x+h)-f(x)=u(x+h)\cdot v(x+h) - u(x+h)\cdot v(x) + u(x+h)\cdot v(x) - u(x)\cdot v(x)= u(x+h)\cdot[v(x+h)- v(x)]+v(x)\cdot [u(x+h)-u(x)][/itex].
    Now divide by h and go to the limit: [itex] f'(x)= \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\rightarrow 0}\frac{u(x+h)\cdot[v(x+h)- v(x)]}{h}+\lim_{h\rightarrow 0}\frac{v(x)\cdot [u(x+h)-u(x)]}{h}=u(x)\cdot v'(x)+u'(x)v(x)[/itex].
     
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