1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper

    I think they want you to do it from the definition. I.e. show that the limit of
    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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Differentiablity proof
  1. A proof. (Replies: 2)

  2. Tips for proofs (Replies: 11)

  3. Simple proof (Replies: 8)

  4. Inequality Proof (Replies: 3)

  5. Proof help? (Replies: 4)