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Derivative of the inverse of a function

  1. Mar 14, 2012 #1
    1. The problem statement, all variables and given/known data
    y=x+(1/x) at y=17/4

    2. Relevant equations

    3. The attempt at a solution
    y^-1: x=y+(1/y)

    differentiate: 1=y'+ln(y)y'
    1=y'(1+ln(y))
    y'=1/(1+ln(y))

    put that over 1: 1+ln(y)

    plug in y: 1+ln(17/4)
    =approximately 2.447
     
    Last edited: Mar 15, 2012
  2. jcsd
  3. Mar 15, 2012 #2

    Mark44

    Staff: Mentor

    Here, by switching letters, you have x = f(y). f is the same function as above, except that its argument is now called y.
    You differentiated with respect to x, so what you have above is 1 = f'(y)*dy/dx
    Now you have dy/dx = 1/f'(y)
    What you're saying and what you're doing are two different things. If you put anything over 1, you get exactly the same thing.

    What you did is take the reciprocal. Why?
    You switched letters, so what was the old y value (17/4) is now an x value. The question is, what is the new y value?

    The answer I get is ~.42
     
  4. Mar 15, 2012 #3

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The inverse function is NOT given by x = y + 1/y

    To see this, just use a test value. e.g., if x = 2, then y(2) will give you some value. If you then plug this value into your "inverse" function (i.e. x(y(2)), you should get back 2. But you don't, because:

    y(2) = 2 + 1/2 = 5/2

    x(5/2) = 5/2 + 2/5 = 25/10 + 4/10 = 29/10

    29/10 is not equal to 2. Since you didn't get back what you started with, your "inverse" function x(y) must be wrong.

    To get the actual inverse function, solve the equation to find x in terms of y. In other words, solve the equation for x. Hint: you'll find that the function does not have a unique inverse.
     
  5. Mar 15, 2012 #4
    42!!!
     
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