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Math Help - Inverse derviative

  1. #1
    Senior Member polymerase's Avatar
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    Inverse derviative

    f(x)=2^{x+1}-2^{-x}. If g(x)=f^{-1}(x) then what is the value of g'(1)? Im telling my friend my she made a mistake...but she won't belive me. I just need a 2nd opinion

    thanks!
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  2. #2
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    Edit: Sorry, this method doesn't work. Kudos to kalagota for figuring out why.

    g'(1) = 1/f'(1)

    f'(x) = 2log(2)2^x+log(2)2^{-x}
    f'(1) = 4log2+.5log(2)=4.5log(2)
    g'(1) = \frac{1}{4.5log(2)}

    There are other ways of expressing this answer, so if you got something different you are not necessarily wrong (assuming I'm right that is).
    Last edited by badgerigar; December 9th 2007 at 02:32 AM. Reason: method was incorrect
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  3. #3
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by badgerigar View Post
    g'(1) = 1/f'(1)

    f'(x) = 2log(2)2^x+log(2)2^{-x}
    f'(1) = 4log2+.5log(2)=4.5log(2)
    g'(1) = \frac{1}{4.5log(2)}

    There are other ways of expressing this answer, so if you got something different you are not necessarily wrong (assuming I'm right that is).
    if g(x) is said to be the inverse of f(x), then it is not necessary that g(x) = \frac{1}{f(x)} (have you tried graphing them?)

    do not confuse the inverse f^{-1}(x) with the power (f(x))^{-1} = \frac{1}{f(x)} they are different..
    Last edited by kalagota; December 8th 2007 at 01:03 AM.
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  4. #4
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    g = f^{-1}

    therefore g(f(x)) = x
    let f(x) = y

    g(y) = x

    differentiating both sides with respect to x and using the chain rule
    <br />
\frac {dy}{dx} \frac{dg}{dy} = 1

     \frac{dg}{dy} = \frac{1}{\frac{dy}{dx}}

    so the derivative of the inverse is the reciprocal of the derivative or I have made a mistake somewhere.
    Last edited by badgerigar; December 8th 2007 at 02:15 AM. Reason: forgot math tags
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  5. #5
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by polymerase View Post
    f(x)=2^{x+1}-2^{-x}. If g(x)=f^{-1}(x) then what is the value of g'(1)? Im telling my friend my she made a mistake...but she won't belive me. I just need a 2nd opinion

    thanks!
    well, i'll do it this way..

    if y = 2^{x+1}-2^{-x}, then the inverse is of the form x = 2^{y+1}-2^{-y} (that is interchange x and y then solve for y in terms of x.. but i'll not do it..)

    if we take the derivative of the latter equation (the inverse) implicitly, we would get

    1 = 2^{y+1} \ln 2 \cdot \frac{dy}{dx} - 2^{-y} \ln 2 \cdot \left( - \frac{dy}{dx} \right) = 2^{y+1} \ln 2 \cdot \frac{dy}{dx} + 2^{-y} \ln 2 \cdot \left( \frac{dy}{dx} \right)

    factoring and isolating \frac{dy}{dx}, we would have

    \frac{dy}{dx} = \dfrac{1}{2^{y+1} \ln 2 + 2^{-y} \ln 2} = \dfrac {1}{\ln 2} \cdot \dfrac{1}{2^{y+1} + 2^{-y}}

    so from x = 2^{y+1}-2^{-y}, if x=1, then

    1 = 2^{y+1}-2^{-y} = \frac{2^{2y+1} -1}{2^y} \implies 2^y = 2\cdot 2^{2y} - 1, so if we set u=2^y, we would get a quadratic equation of the form 2u^2 - u - 1 =0 which factors (2u + 1)(u - 1) = 0 or u = 1 or u = -\frac{1}{2} ; but the second one cannot occur since exponential is always positive.. therefore, u = 1 implying y = 0..

    going back to \frac{dy}{dx} = \dfrac {1}{\ln 2} \cdot \dfrac{1}{2^{y+1} + 2^{-y}} evaluated at y = 0, we have

     <br />
 \dfrac {1}{\ln 2} \cdot \dfrac{1}{2^{1} + 2^{0}} = \frac{1}{3 \ln 2} \approx 0.480898<br />

    too long solution but it is correct..

    look at the attachment..

    the pink one is the fucntion f(x)..
    solid red is the inverse g(x)
    dashed red is the tangent line at x=0
    the blue-green is the identity function y=x..
    Attached Thumbnails Attached Thumbnails Inverse derviative-graph2.jpg  
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  6. #6
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by badgerigar View Post
    g = f^{-1}

    therefore g(f(x)) = x
    let f(x) = y

    g(y) = x

    differentiating both sides with respect to x and using the chain rule
    <br />
\frac {dy}{dx} \frac{dg}{dy} = 1

     \frac{dg}{dy} = \frac{1}{\frac{dy}{dx}}

    so the derivative of the inverse is the reciprocal of the derivative or I have made a mistake somewhere.

    it is correct.. but what is the base of your log? 10? i dont get the same answer..
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  7. #7
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    I was using log to the base e, what you are calling Ln. I don't get the same answer either and just checked your work and couldn't see a mistake.
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  8. #8
    MHF Contributor kalagota's Avatar
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    ohh..

    pardon me.. i actually found where was your mistake..

    yes it is true that g(f(x)) = x

    if you set y=f(x), then g(y)=x..

    the mistake was when you took the derivative with respect to x.. (when you used your chain rule..)

    that should be, \frac{d}{dx}\left( g(y) \right) \cdot \frac{dy}{dx} = 1

    and it will give you \frac{dy}{dx} = f'(x) = 1?

    maybe this is a nicer approach..

    f(g(x)) = x

    thus, f'(g(x)) \cdot g'(x) = 1 \implies g'(x) = \frac{1}{f'(g(x))}..

    or g(f(x)) = x

    and so, g'(f(x)) \cdot f'(x) = 1


    in both cases, you would have to find what g(x) = f^{-1}(x) looks like.. (then, not so nice again.. )
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  9. #9
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    Quote Originally Posted by kalagota View Post
     <br />
\dfrac {1}{\ln 2} \cdot \dfrac{1}{2^{1} + 2^{0}} = \frac{1}{3 \ln 2} \approx 0.480898<br />

    too long solution but it is correct..
    Yes your solution is correct, but we can get it in a direct way:
    Attached Thumbnails Attached Thumbnails Inverse derviative-derivative-inverse.gif  
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  10. #10
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    Here are the graghs of f(x) and its inverse.
    Attached Thumbnails Attached Thumbnails Inverse derviative-derivative-inverse.gif  
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