Results 1 to 5 of 5

Math Help - derivative of an invertible function

  1. #1
    Member
    Joined
    Sep 2005
    Posts
    136

    derivative of an invertible function

    If f(x) is invertible with inverse g(y) then g ' (y) is
    (a) 1/f(g(y))
    (b)1/ f ' (g))
    (c)1/f(x)
    (d) undeined
    (e) none of the above
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,855
    Thanks
    321
    Awards
    1
    Quote Originally Posted by bobby77 View Post
    If f(x) is invertible with inverse g(y) then g ' (y) is
    (a) 1/f(g(y))
    (b)1/ f ' (g))
    (c)1/f(x)
    (d) undeined
    (e) none of the above
    The inverse function theorem says that if y = f(x) has an inverse x = g(y), then then g'(y) will be 1/f'(x). There is an easy way to see this (though it is much more difficult to prove!):

    If we have a function y(x), then dx/dy = 1/(dy/dx). You can "clear the fraction" in the "denominator" and see that the RHS is also dx/dy. By construction of the derivative, dx/dy would be the derivative of the inverse function.

    Before someone more Mathematically minded than me comes along to remind you, let me say that treating dy/dx as a fraction is a very tricky process and is not a strictly legitimate process even where you can get away with it. (But it does occasionally provide a convenient guideline.)

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by topsquark View Post
    Before someone more Mathematically minded than me comes along to remind you, let me say that treating dy/dx as a fraction is a very tricky process and is not a strictly legitimate process even where you can get away with it. (But it does occasionally provide a convenient guideline.)
    I never, ever, never use it. You can determine which class a person belongs to: a low class (physics) if he uses that trick, a high class (math) if he does not.
    ---
    The complete proof is not so easy, it relys on properties of the interval.

    Anyway here is an almost complete proof.

    Let f be a differenciable function continous on the open interval (a,b).
    Assume, f has an inverse differenciable function g.
    If and only if,
    f o g=x on (a,b)
    Then the derivative is (chain rule: both are differenciable),
    g' * f' o g=1
    Note that f' o g is never zero on (a,b).
    Equivalently,
    g' = 1/(f' o g)
    ~~~
    Why is this proof incomplete?
    Actually this proof is complete, I did not prove that if f is differenciable and bijective on (a,b) then so too must be g, I assumed that if that is true then that is the case.

    But that is not really such a bother if you are working with a "well-behave" invertible function then all you need to show that its inverse is differenciable and thus apply this proof.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,855
    Thanks
    321
    Awards
    1
    Quote Originally Posted by ThePerfectHacker View Post
    I never, ever, never use it. You can determine which class a person belongs to: a low class (physics) if he uses that trick, a high class (math) if he does not.
    You can't deny that it's a useful mnemonic, though.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by topsquark View Post
    You can't deny that it's a useful mnemonic, though.
    Not really.
    Again, never use it never will.
    Even when it comes to solving differencial equations.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. when is function F locally invertible
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: May 4th 2009, 10:31 PM
  2. Help on invertible function question
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: November 14th 2008, 02:45 PM
  3. invertible increasing function
    Posted in the Pre-Calculus Forum
    Replies: 0
    Last Post: August 18th 2008, 06:13 AM
  4. Invertible function problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 8th 2007, 12:22 PM
  5. invertible function by graphs
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: February 5th 2007, 06:43 AM

Search Tags


/mathhelpforum @mathhelpforum