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
    10,211
    Thanks
    419
    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
    10
    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
    10,211
    Thanks
    419
    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
    10
    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, 11:31 PM
  2. Help on invertible function question
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: November 14th 2008, 03:45 PM
  3. invertible increasing function
    Posted in the Pre-Calculus Forum
    Replies: 0
    Last Post: August 18th 2008, 07:13 AM
  4. Invertible function problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 8th 2007, 01:22 PM
  5. invertible function by graphs
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: February 5th 2007, 07:43 AM

Search Tags


/mathhelpforum @mathhelpforum