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Homework Help: Derivative homework help

  1. Dec 25, 2011 #1
    1. The problem statement, all variables and given/known data

    [tex]\lim_{h\to 0}\frac{F[p(x)+hp'(x)]-F[p(x)]}{h}[/tex]

    where [tex]F'=f[/tex]

    2. Relevant equations
    [tex]\lim_{h\to 0}\frac{F(x+h)-F(x)}{h}=F'(x)[/tex]

    3. The attempt at a solution

    I think that solution is [tex]p'(x)f[p(x)][/tex] but I have a trouble to get the result.
  2. jcsd
  3. Dec 25, 2011 #2
    Re: Derivative

    What is ##p'(x)##?

    It is

    [tex] \lim_{h \to 0} \frac{p(x+h)-p(x)}{h} [/tex]


    [tex] p(x)+hp'(x)=p(x+h) [/tex]

    So now we have

    [tex] \lim_{h\to 0}\frac{F(p(x+h))-F(p(x))}{h}=F'(p(x))=p'(x)f(p(x))[/tex]
  4. Dec 25, 2011 #3

    Ray Vickson

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    Science Advisor
    Homework Helper

    Re: Derivative

    The first result [tex] p(x)+hp'(x)=p(x+h) [/tex] is incorrect, although it can be fixed up. The last line that writes [tex] \lim_{h\to 0}\frac{F(p(x+h))-F(p(x))}{h}=F'(p(x))[/tex] is also incorrect; it should be [itex] dF(p(x))/dx[/itex], not [itex] F'(p(x))[/itex], because, in fact, [itex] F'(p(x)) = f(p(x).[/itex]

    A much easier approach is: let [tex] R(x,h) = \frac{F[p(x)+hp'(x)]-F[p(x)]}{h}. [/tex]
    (i) if p'(x) = 0 (which can, perhaps, happen at a given point x), then for all [itex] h \neq 0,[/itex] R(x,h) = 0, so the limit is zero, as is [itex] p'(x) f(p(x)).[/itex] (ii) if [itex] p'(x) \neq 0, [/itex], let [itex] k = h p'(x)[/itex], so that
    [tex] R(x,h) = p'(x) \frac{F[p(x)+k)-F[p(x)]}{k}. [/tex] As [itex] h \rightarrow 0[/itex] we have also that [itex] k \rightarrow 0 [/itex], so the limit is [itex] p'(x) f(p(x)).[/itex]

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