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Homework Help: Given f(x) = 4^√x - x + 3, find (f^-1)'(3).

  1. Jul 22, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the value of the derivative of the inverse of the function.

    Given that f(x) = (4^√x) - x + 3, find (f^-1)'(3).



    2. Relevant equations



    3. The attempt at a solution

    I know that to find the inverse derivative, one should switch the x and y's and then take the derivative, but that 4^√x is kind of throwing me off...
     
  2. jcsd
  3. Jul 22, 2011 #2

    Pyrrhus

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    You just need to compute [itex]\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} [/itex]

    You specify z= (4^√x), and then calculate dz/dx. You can do that by applying natural log as ln z = √x ln 4, take derivative, and then go back to compute the original dz/dx.

    You can do this because if h(t) = u(t) + w(t) and you know the differential operator has the linear property, and thus h' = u' + w', where the primes mean derivatives wrt to t.
     
    Last edited: Jul 22, 2011
  4. Jul 22, 2011 #3
    Pyrrhus, is right, but from experience many introductory calculus students won't understand everything he's saying.


    http://tutorial.math.lamar.edu/Classes/CalcI/DiffInvTrigFcns.aspx" [Broken]
    Keep in mind that site uses f(x)=f(x) sometimes and g(x) = [itex]f^{-1} (x)[/itex] other times.
    So context is essential for knowing if x is the horizontal, independent variable or x might better be thought of as the vertical, dependent variable. Alternately, the notation and concepts may be easier to understand if we ignore the usual interpretation of x vs. y.

    To evaluate g'(3) = [itex]f^{-1}[/itex]' we'll need to first find
    g(3). We need to apply f( g(3) ) = 3 to find g(3).
    So f(x) = 3.
    I'll let you work out x. For now, I'll call it b.
    Next step, find f'. You'll need.

    d/dx [itex]4^{\sqrt{x}}[/itex]
    First look up d/dt [itex]a^{t}[/itex] in a derivative table.
    Then apply chain rule.

    I trust you can find
    g'(3) = [itex]\frac{1}{f'(b)}[/itex]
    from there.

    Show us what that gives you or if you get the answer in the back of the book.
     
    Last edited by a moderator: May 5, 2017
  5. Jul 22, 2011 #4

    hunt_mat

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    In general though let [itex]y=f^{-1}(x)[/itex], apply the function f to both sides to obtain [itex]f(y)=x[/itex], differentiate with respect to x to obtain:
    [tex]
    f'(y)\frac{dy}{dx}=1
    [/tex]
    Divide by f'(y) to obtain:
    [tex]
    \frac{dy}{dx}=\frac{1}{f(y)}=\frac{1}{f(f^{-1}(x))}
    [/tex]
    With your problem, swap f and the inverse of f around, so [itex]y=f(x)[/itex] and [itex]x=f^{-1}(y)[/itex]
     
    Last edited: Jul 22, 2011
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