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Differentiablity proof

  1. Nov 12, 2009 #1
    Suppose that a function f is differentiable at x0 and define g(x)=f(mx+b), where m and b are constants. Prove that if x1 is a point at which mx1+b=x0, then g(x) is differentiable at x1 and g'(x1)=mf'(x0).
    Definition: A function f is said to be "differentiable at x0" if the limit
    [tex] f'(x_{0})= lim_{h \rightarrow 0} \frac{f(x_{0}+h)-f(x_{0})}{h}[/tex]
    exists. If f is differentiable at each point of the open interval(a,b), then we say that f is "differentiable on (a,b)", and similarily for open intervals of the form [tex] (a,+\infty),(-\infty,b)and (-\infty,+\infty)[/tex]. In the last case we say f is "differentiable everywhere". I think this is a simple proof, but since I am doing this no my own with a book that doesn't do too many example proofs, I am lost. g(x1) = f(mx1+b) = f(x0), which I think proves the limit of g(x1) exists but g'(x1)=? Thanks for the help.
     
  2. jcsd
  3. Nov 13, 2009 #2

    CompuChip

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    I think they want you to do it from the definition. I.e. show that the limit of
    [tex]
    \frac{g(x_{1}+h)-g(x_{1})}{h}
    [/tex]
    for h to zero exists by substituting the definition of g and using that the appropriate limit for f exists.
     
  4. Nov 13, 2009 #3
    Sorry, if I gave the wrong impression, but the definition I included myself thinking it relevant. The definition was not part of the original question.
     
  5. Nov 14, 2009 #4

    CompuChip

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    If you don't want to use the definition, you can directly prove if from the chain rule, I suppose.

    I checked the proof from the definition, it works out nicely. Just remember to bring the limit you get back in the form
    [tex]\lim_{\delta \to 0} \frac{f(x_0 + \delta) - f(x_0)}{\delta} [/tex]
    where [itex]\delta[/itex] is some new quantity depending on h.

    (I cannot really give you more hints without doing the calculation for you, so I suggest you give this a try and post what you get).
     
  6. Nov 14, 2009 #5
    [tex] \lim_{h \rightarrow 0} \frac{ g(x_{1} +h) - g(x_{1})}{h} = \lim_{h \rightarrow 0} \frac{f(m(x_{1}+h)+b) - f(mx_{1} + b)}{h} [/tex]
    [tex] = \lim_{h \rightarrow 0} \frac{ f(mx_{1} + mh + b) - f(mx_{1} + b)}{h}[/tex]
    [tex] = \lim_{mh \rightarrow 0} m\frac{ f(mx_{1}+mh+b)-f(mx_{1}+b)}{mh}[/tex]
    [tex] = \lim_{mh \rightarrow 0} m\frac{ f(mx_{1}+b+mh)-f(mx_{1}+b)}{mh}[/tex]
    [tex] = m \lim_{\delta \rightarrow 0} \frac{ f(x_{0}+ \delta)-f(x_{0})}{\delta}[/tex]
    [tex] = mf'(x_{0})[/tex]

    If this is ok, now I got to use this result to prove that [tex] f'(x)=nm(mx+b)^{n-1} [/tex] if [tex] f(x) = (mx+b)^{n}[/tex], where m and b are constants and n is any integer. I know that [tex] \frac{dx^n}{dx}=nx^{n-1} [/tex] where n is any integer or do I use the definition given above in #1? Thanks again.
     
    Last edited: Nov 14, 2009
  7. Nov 15, 2009 #6

    CompuChip

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    Very well.

    The quickest way indeed seems to be, to use that
    [tex]\frac{d}{dx}f(x) = n x^{n-1}[/tex]
    where f(x) = xn is differentiable anywhere, and use what you have just proven.
     
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