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Derivative of inverse function at x=0

  1. Mar 3, 2012 #1
    Derivative of inverse function at x=0 [SOLVED]

    1. The problem statement, all variables and given/known data

    Let f(x) = x + Ln(x+1), x > -1

    Find [itex]\frac{d}{dx}[/itex] [itex]f^{-1}[/itex] [itex]|_{x=0}[/itex]; Note that f(0) = 0

    2. Relevant equations

    [itex](f^{-1})'[/itex](x) = [itex]\frac{1}{f^{'}(f^{-1}(x))}[/itex]

    (or)

    [itex]\frac{dx}{dy}[/itex] = [itex]1/\frac{dy}{dx}[/itex]

    3. The attempt at a solution

    I attempted to find the inverse of f(x) by switching x & y and solving for y. This wouldn't work so I assume the function has no inverse.

    I've read that if [itex]f^{-1}[/itex] is differentiable at x, the slope of the tangent line will be given by the value of the derivative of [itex]f^{-1}[/itex]at x. Since the graph of [itex]f^{-1}[/itex] is obtained from the graph of f by reflection across y=x, the same is true for the tangent lines.

    Then, would finding the derivative of f(x=0),

    f'(x) = 1 + ([itex]\frac{1}{x+1}[/itex])(1)
    f'(x=0) = 1 + [itex]\frac{1}{1+0}[/itex] = 2

    and then flipping it for its inverse [itex]\frac{1}{2}[/itex] be a legitimate means of finding the first derivative of the inverse of f(x) at point x=0 (even if I'm not sure the inverse exists)?
     
    Last edited: Mar 4, 2012
  2. jcsd
  3. Mar 3, 2012 #2

    Dick

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    Ok then, you only have to find f^(-1)(0). You don't need to find f^(-1)(x) for any x. What's f^(-1)(0)?
     
  4. Mar 3, 2012 #3

    Dick

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    Ok, you've got it. I'm pretty sure the inverse does exist. f'(x)>0 for x>(-1).
     
  5. Mar 3, 2012 #4
    If you mean I just find

    f(x=0) = 0 + ln(1)

    and take its inverse,

    1/0 DNE

    and I don't get anywhere.

    The more I look at it I believe that the inverse just DNE and this problem is a trap.
     
  6. Mar 3, 2012 #5
    Ok, so the way I did it seems to be the only possible way to arrive at a solution? I worried about it, because I've tried the method in a test problem that I could do normally with no success:

    Let f(x)= 2x^2-2, x is greater than or equal to zero.

    Find d/dx f-inverse(x)|x=0 [Note than f(1)=0]

    Answer: d/dx f-inverse(0)= 1/4

    f'(x) = 4x
    f'(x=0) = 0
    flip for inverse 1/0
    (instead of 1/4 as I know the answer to be)
     
  7. Mar 3, 2012 #6

    Dick

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    No, f^(-1)(0) is supposed to be the inverse function, not 1/f(0). f^(-1)(0)=0. You just did it correctly, why are you retracting that?
     
  8. Mar 3, 2012 #7

    Dick

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    f^(-1)(0)=1. f'(1)=4. 1/f'(f^(-1)(0))=1/f'(1)=1/4. So I'm not sure what is worrying you.
     
  9. Mar 3, 2012 #8
    I see. I'm just having trouble finding an inverse for the original problem, and its throwing me off.

    Wolfram gave me an inverse using the product log function, which I know isn't expected in this course.

    Can I ask how you verified f^(-1)(0)=0? I understand everything else you've said and can finish from there.
     
  10. Mar 3, 2012 #9

    Dick

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    I'm assuming you switched back to the first problem. If f(x)=x + ln(x+1), then f(0)=0, so f^(-1)(0)=0, right?
     
  11. Mar 3, 2012 #10
    Yes, original problem. I see, you're finding the reflection of the point f(0)=0 across y=x (so the point (0,0)). I was making the mistake of trying to use a multiplicative inverse. I think I've got it now, thank you.
     
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