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Derivative of inverse function.

  1. Oct 28, 2011 #1
    1. The problem statement, all variables and given/known data
    Let, f(x) = x^3 - 3x^2 - 1 , x => 2 . Find the value of df^-1/dx at the point x=-1=f(3)

    2. Relevant equations
    The definition states; "If f has an interval I as domain and f'(x) exists and is never zero on I, then f'(c) is diff at every point in its domain. The value of (f^-1)' at a point b in the domain of f_1 is the reciprocal of the value of f1 at the point a = f^-1(b):
    Though I don't really understand what is going on in the equation.

    3. The attempt at a solution
    The derivative is f' = 3x^2 - 6x , but I don't really get what and why I do what I have to do from here. Coulden't find a youtube video for these sort of problems.
  2. jcsd
  3. Oct 28, 2011 #2

    Char. Limit

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    Gold Member

    Wow, that definition sounds complicated. The definition I always used is, well... y = f(x), yeah? So the inverse of f(x) would be x(y), and the derivative of that is dx/dy. Now it might seem intuitive, and there are pitfalls in that, but this is true because of the chain rule:

    [tex]\left(f^{-1}(x)\right)' = \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}[/tex]

    So the derivative of the inverse of your function is the reciprocal of the derivative of your function. Fun stuff, really.
  4. Oct 28, 2011 #3
    Oh, ok. So, could you tell me what does the question ask for when they ask to find df^-1/dx at x=-1=f(3) ?
  5. Oct 28, 2011 #4


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    Science Advisor

    it makes more sense if you write it this way:

    dx/dy = 1/(dy/dx), but this isn't very accurate notation.

    in english: the derivative of the inverse, is the reciprocal of the derivative.

    in other words, instead of writing y = f(x), you consider x = f-1(y).

    your equation should read:

    [tex](f^{-1})'(f(a)) = \frac{1}{f'(a)}[/tex] or

    [tex](f^{-1})'(b) = \frac{1}{f'(f^{-1}(b))}[/tex]

    since presumably b = f(a) (and thus a = f-1(b)).
  6. Oct 28, 2011 #5
    So, in this case I don't have to find the inverse of the function because x=-1=f(3);
    [tex](f^{-1})'(b) = \frac{1}{f'(f^{-1}(3))}[/tex] ----> [tex](f^{-1})'(b) = \frac{1}{f'(-1)}[/tex] ?
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