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Calculate second derivative of a function in terms of another function

  1. Dec 14, 2013 #1
    The problem statement, all variables and given/known data.

    Let ##f:[a,b] \to [\alpha,\beta]## be a bijective function of class ##C^2## with an inverse function also of class ##C^2## ##g:[\alpha,\beta] \to [a,b]##.

    a)Calculate ##g''(x)## for every ##x \in (\alpha,\beta)## in terms of ##f## and its derivatives.
    b)If ##f'(x)>0## and ##f''(x)>0## for every ##x \in (a,b)##, prove that ##g''(x)<0## for every ##x \in (\alpha,\beta)##

    For part ##a)##, all I could do was:

    ##g(y)=(g\circ f)(x)##, so ##g'(y)=(g\circ f)'(x)=g'(f(x))f'(x)##.
    Then, ##g''(y)=\dfrac{d}{dx}g'(y)=\dfrac{d}{dx}g'(f(x))f'(x)##.

    Computing this derivative gives:
    ##\dfrac{d}{dx}g'(f(x)).f'(x)=g''(f(x){f'(x)}^2+g'(f(x))f''(x)##

    Now, I need to find ##g''(y)## just in terms of ##f## and its derivatives, I don't know how to get rid of ##g''(f(x))## and ##g'(f(x))##.

    For part ##b)## I have no idea what to do, I would appreciate any suggestion or idea to prove ##b)##.
     
    Last edited: Dec 14, 2013
  2. jcsd
  3. Dec 14, 2013 #2
    What you are missing is that g(f(x)) = x. Plug that into your first equation, and you will find things simplify out nicely.

    What you computed so far is just d/dx(f(g(x)) without using the fact that these are inverses.

    Part b is about the nature of inverses. For example, if f is continuous, so is its inverse; if discontinuous so is its inverse, if differentiable once, so is its inverse, etc.

    Here's how to do some thinking about things:

    What if f is monotonic increasing? What could you say about f inverse? Take some examples -- look at f(x) = ##x^2## on [0,1]. What is its inverse there? Does f increase or decrease ? What about its inverse? Based on this one example would you feel safe generalizing? Or would you feel safe after looking at 2 or 3 examples? I realize anything can happen on example 4, but when you look at these examples, you can probably see a reason why things are working out as they do; and that reason would apply to all invertible functions.

    Once you get that far, look at your computation for g'(x) in terms of f. Does that bear out the intuitive conclusions you drew above?

    Okay, go on to the f'' case. In the example above what happens to g''? Why?

    Now you can construct an argument about f'' > 0 implying g'' < 0. You might use your computation in part a. Or you might imagine a geometric argument about f inverse being the reflection of f across the line y = x.
     
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