1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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]


    [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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook