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Derivative inverse function

  1. Jun 21, 2011 #1
    Hello,

    I'am suffering with the theoretical background. My course state the follow thing:

    D(f-1(y))=1/D(f(x)).

    So: f-1(y) is the inverse function of f(x), this means that the argument of f-1(y) is y!

    Example: y = f(x) = x^2 => f-1(y): x = y^2. Am I correct with is one ?

    The chain rule is applied at the following equation: D(f-1(y))=D(f-1(f(x)))=
    (f-1(y))'.(f(x))'=1, this is because (f-1(f(x)) = x.

    I do understand this, but I'm very confused with the next on:
    (f-1(y))'=d(f-1(y))/dy, this is because the inverse function has y as argument.
    Applied at my example I would get: d(f-1(y))/dy = 2.y

    And by D(f-1(y))=1/D(f(x)), the derivative of the inverse function would be: 1/(d(x^2)/dx)
    and this is 2.x

    I TOTALLY don't understand it..

    Someone dissolve my confussion ?

    greetz
     
  2. jcsd
  3. Jun 21, 2011 #2

    hunt_mat

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    Homework Helper

    Let [itex]y=f^{-1}(x)[/itex], then [itex]x=f(y)[/itex]. differentiating this shows that: [itex]1=f'(y)dy/dx[/itex] then:
    [tex]
    \frac{dy}{dx}=\frac{1}{f'(y)}=\frac{1}{f'(f^{-1}(x))}
    [/tex]
     
  4. Jun 21, 2011 #3
    No, x = sqrt(y) or x = -sqrt(y)
    .
     
  5. Jun 21, 2011 #4
    @hunt mat: can you explain that to me with an example ?

    @JJacquelin:
    I made a graph of what you wrote:
    inverse%25function.png

    y=x^2 is the explicit function (purple)
    x-sqrt(y) = 0 is the implicit function (red)
    those are the same functions...

    I think the inverse function is: f-1(y)=y^2 =x
    because f-1(x)=sqrt(x)
    look at this graph:

    inverse%252.png

    greetz
     
  6. Jun 21, 2011 #5

    hunt_mat

    User Avatar
    Homework Helper

    calculate the derivative of [itex]y=\sin^{-1}x[/itex], take sin of both sides to find x=sin y, differentiate this to find:
    [tex]
    1=\cos y\frac{dy}{dx}
    [/tex]
    Then:
    [tex]
    \frac{dy}{dx}=\frac{1}{\cos y}
    [/tex]
    Upon using [itex]\sin^{2}y+\cos^{2}y=1[/itex], this shows that [itex]\cos y=\sqrt{1-\sin^{2}y}=\sqrt{1-x^{2}}[/itex]. Inserting this into the equation we can see that:
    [tex]
    \frac{dy}{dx}=\frac{1}{\sqrt{1-x^{2}}}
    [/tex]
     
  7. Jun 21, 2011 #6
    nietschje, I think you're confusing yourself with what you're using y for. Forget what the function f takes as an argument for now--it's just the argument, and you can call its input whatever you want.

    When you find the inverse of a function, it's of the function--the arguments just provide a notationally easy way to understand what you're doing and compute what f^-1, the inverse of f, should be.

    So your situation is, you have the function f(x) = x^2. Then to find the inverse of f, you're defining a new variable, y = f(x), that is a function of x.

    Now to find the inverse of f (note that it's not unique, depending on what the function f is, but to find the inverse for at least part of the range of f...), you're finding x in terms of y, instead of what you start with, which is y in terms of x.

    So if y = f(x) = x^2, we have:

    y = x^2 = f(x)

    so then it must be that:

    sqrt(y) = x

    So f^-1(y) = sqrt(y) is your inverse function -- if you want to continue to think of all this in terms of the argument the functions take, just remember that the inverse function, in your notation, takes y's and goes to x's -- where f took x's to y's. So the inverse function is f^-1 : y --> sqrt(y).

    That's the same as f^-1(x)=sqrt(x). It's just renaming the variable you used.


    One last note--you wrote "f^-1(x)=sqrt(x) " and concluded that f^-1(y) = y^2. Remember, f^-1 is the same function no matter what you call the input. So if f^-1(x) = sqrt(x), as you claim (correctly), then you have to have f^-1(y) = sqrt(y). (since you appear to be taking x and y to be general variables over the whole domain of f^-1, the same domain in each case).
     
  8. Jun 22, 2011 #7
    Hello nietschje !

    you make a confusion between different thinks :
    First the notation (x,y) of function y=f(x)
    Second, the notation of the axes of the plot. Use different notations, for exemple (X,Y) instead of (x,y). So there will be no confusion between (x,y) and (X,Y) which correspond to different objects : the first is a function, the second is a system of axes.
    Then, you want to draw the function y=f(x)=x². So, you decide that X=x and Y=y . The curve obtained corresponding to Y=X² is a parabola.
    Now, the inverse function is x=(+or-)sqrt(y). You want to draw it. Now x is a function of y, so Y=x and X=y. The curve obtained corresponds to Y=(+or-)sqrt(X) . This curve is also a parabola, but with horizontal axe instead of vertical.
    Always, the curve (C1) which represents a function and the curve (C2) which represents the inverse function, are symetric relatively to the straight line Y=X. Do not expect that (C1) and (C2) be se same curve because the function and the inverse function are not the same function : the first is y=x² where y is a function of x ; The second is x=(+or-)sqrt(y) where x is a function of y. They are different since power 2 is different from square root.
     
  9. Jun 22, 2011 #8
    I read all comments, thanks for your quick replies!

    @krb17:
    for my example:
    y = x^2 ; given x = +(-)2 => y (looking for) = 4

    x = +(-)sqrt(y) ; given y = 4 => x (looking for) = +(-) 2
    The purpose of the inverse function was that we want to find the value of x when the value of y was given, according to the function of f(x)

    y = x^2 and x = +(-)sqrt(y) are both paraboles and the same functions ! This is because there arent changes in de x y axles ! See my example...

    I suppose I am correct (cant find any mistakes :) )

    Buth when I read that the definition of the inverse function its states that the function y = f(x) is reflected accros the function y = f(x) = x.
    I know that I can transform x =+(-)sqrt(y) to y = +(-)sqrt(x) and
    this is graphiccaly the inverse funtion of x = +(-)sqrt(y) !
    Because the argument x (x axles) is literally changed (rotate this graph 90° counterclockwise and you see a parabole)...

    Alright but now its get very confussing to me:
    f-1(y) = +(-)sqrt(y) = x : I do understand this is called the inverse function because of my example( see above) but graphiccaly it doesn't make any sense...
    f-1(x)=+(-)sqrt(x) = y : I can SEE that this is the inverse function graphiccaly but using number its nog making any sense
    I am very confussed with this because of the contradictory definitions...

    In terms of derivating: d(f-1(y)/dy = 1/d(f(x)/dx

    But f-1(y) isn't graphiccaly the inverse funtion (prooved it above) so must I interprete this as: tranform y = f(x) to x = f(y) still same function! And derivative f(y) to y... ?
    Because graphiccaly there are not any changes and the derivatives is graphiccaly the tangent value...


    Sorry for the long text ! Hope I made clear where my confussing comes from... Hard time to understand this concept...

    greetz
     
  10. Jun 22, 2011 #9
    Absurd !
    if y=f(x) then x=g(y) where function g is different from function f.
    This is written in my preceeding post. Citation : "the function and the inverse function are not the same function".
    Moreover you refuse to use different notations for functions (x,y) and axes (X,Y), so your last post is full of mix-up.
    Enough explanation and example was given. Make the effort to carefully read the explanations and take enough time to think at each sentence of the explanations. That is well sufficient and I will no longer answer to more ill-considered questions.
     
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