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Homework Help: Derivative product rule and other rule proofs.

  1. Mar 20, 2012 #1
    1. The problem statement, all variables and given/known data

    Prove that the functions: (u+v)'(x0) and αu and u*v are derivable.

    2. Relevant equations
    in other words prove that :

    [tex](\alpha u)'(x_{0})=\alpha u'(x_{0})[/tex]

    [tex](u\cdot v)'(x_{0})=u'(x_{0})\cdot v(x_{0})+u(x_{0})\cdot v'(x_{0})[/tex]

    3. The attempt at a solution

    Can someone give me some heads up on how to start this proof and should i use the limit of (f(x)-f(x0))/x-x0 to prove this? We have never done f'(x)=lim as h approaches 0 of (f(x+h)-f(x))/h
    Last edited: Mar 20, 2012
  2. jcsd
  3. Mar 20, 2012 #2


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    Homework Helper

    yes, first principles is the way to go with all of these, here's some tex to write it if that helps for future (right click to see the code)
    f'(x) = \lim_{h \to 0}\frac{f(x+h)-f(x)}{h}
  4. Mar 20, 2012 #3


    Staff: Mentor

    Yes, it looks to me like you need to use the definition of the derivative for all three parts.
  5. Mar 20, 2012 #4
    yes but we never did derivatives with f(x+h)-f(h)/h
  6. Mar 20, 2012 #5


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    Staff Emeritus
    Science Advisor
    Gold Member

    You realize that these two things in bold are equivalent, right? The interval "h" is just the difference between the two values x and x0 where you are evaluating the function. So saying that h → 0, is the same as saying that x → x0. So if you've seen the former, then you know how to solve the problem.
  7. Mar 20, 2012 #6
    Ok i've proved all of them and the one with alpha is proved in one line i don't know if it's that easy, but that's what i got.
  8. Mar 20, 2012 #7


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    Homework Helper

    should be pretty straight forward once you get the definition down
    Last edited: Mar 20, 2012
  9. Mar 20, 2012 #8
    Yes thank you very much i've done all of them and by the way proving the quotient rule and (1/v)(x0) should be the same right?
  10. Mar 20, 2012 #9


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    Homework Helper

    will need a little more manipulation, but the same approach will work
  11. Mar 20, 2012 #10
    Yes thank you I've just proven all of the stuff i had to do.
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