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Product rule in differentiation

  1. May 8, 2015 #1
    What has done here in the second line of the proof for product rule?, from Mathematical methods for physicists from Riley, Hobson
    they defined f(x)=u(x)v(x) and these steps are given,
    I have no idea how to proceed further please help me.
     

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  3. May 8, 2015 #2

    Simon Bridge

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    Do you know why they bothered to work out ##f(x+\Delta x) - f(x)## at all?
    I mean: what's the point?

    How would you normally go about proving the product rule - if you didn't have the example from Riley and Hobson?
    i.e. do you know the definition of the derivative?
     
  4. May 9, 2015 #3
    I don't know

    yes, derivative is the rate of one function to another function, it actually says how fast one function changes with respect to other, am I right?
     
  5. May 9, 2015 #4

    Simon Bridge

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    Not exactly ... that was the description of what the derivative is, not the definition. The definition is: $$f^\prime(x) = \lim_{\Delta x\to 0}\frac{f(x+\Delta x) - f(x)}{\Delta x}$$ ... this gives the derivative of f(x) with respect to x.

    To prove the product rule, first set ##f(x)=v(x)u(x)## then apply the definition to f.
     
  6. May 9, 2015 #5
    got it. Thank you
     
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