Results 1 to 10 of 10

Math Help - Stewart's Proof that Inverses of Differentiable Functions are Differentiable

  1. #1
    Junior Member
    Joined
    Oct 2008
    Posts
    41

    Stewart's Proof that Inverses of Differentiable Functions are Differentiable

    The following is paraphrased from Stewart's Calculus text.

    Theorem: If f is a one-to-one differentiable function with inverse function g and f'(g(a)) \ne 0, then the inverse function is differentiable at a and g'(a)=\frac{1}{f'(g(a))}.

    Proof: We have g'(a)=\displaystyle{\lim_{x\rightarrow a}\frac{g(x)-g(a)}{x-a}}. Since f and g are inverses we have g(x)=y iff f(y)=x and g(a)=b iff f(b)=a.

    Since f is differentiable, it is continuous, so g is continuous by earlier theorem. Thus as x\rightarrow a we have g(x)\rightarrow g(a), or y\rightarrow b. Therefore:

    g'(a)=\displaystyle{\lim_{x\rightarrow a}\frac{g(x)-g(a)}{x-a}}

    ** =\displaystyle{\lim_{y\rightarrow b}\frac{y-b}{f(y)-f(b)}}

    ...

    =\frac{1}{f'(g(a))}

    The starred step is the one I'm having difficulty with understanding fully. I understand the continuity argument that as x\rightarrow a we have y\rightarrow b, but it seems like a bit of a leap, at least I can't think of any definitions or theorems to apply, to use that to change \lim_{x\rightarrow a} to \lim_{y\rightarrow b} (although an \epsilon - \delta argument seems to work in the opposite direction--that \lim_{y\rightarrow b}h(x)=L would imply that \lim_{x\rightarrow a}h(x)=L). It seems that the claim in Stewart's argument also has to do with the fact that f and g are one-to-one. Any thoughts?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,663
    Thanks
    1616
    Awards
    1
    Quote Originally Posted by Diamondlance View Post
    Since f is differentiable, it is continuous, so g is continuous by earlier theorem. Thus as x\rightarrow a we have y\rightarrow b.
    That seems to explain it does it not?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Dec 2010
    Posts
    470
    Think about it this way.

    Let h(x) = \displaystyle{\frac{g(x)-g(a)}{x-a}}.

    So g'(a) = \displaystyle{\lim_{x\rightarrow a}h(x)}, and since f is continuous and has f'(b)\neq 0:

    g'(a) = \displaystyle{\lim_{y\rightarrow b}h(f(y))}

    The rest of the step is just algebraic manipulation.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Oct 2008
    Posts
    41
    Snowtea, that makes sense, and helps to alleviate the problem I was having.

    What I was thinking was that, for example, though x\rightarrow 2 implies x^2\rightarrow 4,

    \lim_{x\rightarrow 2}x=2 whereas \lim_{x^2\rightarrow 4}x doesn't make sense since x is not a function of x^2 on \mathbb{R}. However, since x is a continuous function of y here, the switch is justified.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Oct 2008
    Posts
    41
    So I suppose the more rigorous argument I was searching for would go as follows:

    Given one to one continuous function g such that \displaystyle{\lim_{g(x)\rightarrow g(a)}\frac{g(x)-g(a)}{x-a}} exists and is equal to g'(a), we claim that \displaystyle\lim_{x\rightarrow a}\frac{g(x)-g(a)}{x-a}}=g'(a). Given \epsilon >0, choose \delta_1,\delta_2>0 such that

    0<|x-a|<\delta_1

    implies that 0<|g(x)-g(a)|<\delta_2 by continuity and injectivity of g,

    where \delta_2 is small enough so that

    0<|g(x)-g(a)|<\delta_2 implies that

    |\frac{g(x)-g(a)}{x-a}-g'(a)|<\epsilon by the assumed limit.
    Last edited by Diamondlance; December 16th 2010 at 02:03 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Dec 2010
    Posts
    470
    I don't think it is good notation for the limit to write \displaystyle{\lim_{g(x)\rightarrow g(a)}\frac{g(x)-g(a)}{x-a}}.
    It's better to write \displaystyle{\lim_{y\rightarrow b}\frac{g(f(y))-g(a)}{f(y)-a}}

    we can rewrite
    \displaystyle{\frac{g(f(y))-g(a)}{f(y)-a}} = \frac{y-b}{f(y)-f(b)}}} by definition.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Or we cuould just note that f(g(x))=x and so by the chain rule we get f'(g(x))g'(x)=1. Dividing both sides by f'(g(x)) (since it's non-zero) gives the answer.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Oct 2008
    Posts
    41
    I don't think so, since the differentiability of g(x) is not assumed.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,663
    Thanks
    1616
    Awards
    1
    Quote Originally Posted by Diamondlance View Post
    I don't think so, since the differentiability of g(x) is not assumed.
    Actually that is a well known theorem.
    If f is a one-to-one differentiable function in an open interval \mathcal{I} then f^{-1} exists on \mathcal{I} and is differentiable there.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member
    Joined
    Oct 2008
    Posts
    41
    Absolutely, but that's basically what we're trying to prove.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Differentiable functions
    Posted in the Calculus Forum
    Replies: 7
    Last Post: December 7th 2011, 01:53 AM
  2. A proof about differentiable functions
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 8th 2011, 11:07 PM
  3. Replies: 0
    Last Post: October 3rd 2010, 07:03 AM
  4. Differentiable functions.
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 17th 2009, 02:47 PM
  5. Sequence of differentiable functions, non-differentiable limit
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: April 3rd 2009, 05:13 AM

Search Tags


/mathhelpforum @mathhelpforum