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Inverse Function with Differentiation

  1. Nov 8, 2005 #1
    Let f(x) = x^3 + e^x.
    Find (f^-1)'(2).
    I know how to do everything else except the first step. How do you find the inverse of f(x)? I know the inverse of an exponential function is a logarthmic function, but where do I proceed from here?
    Thanks.
     
  2. jcsd
  3. Nov 8, 2005 #2

    benorin

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    Homework Helper

    There's a theorem for that

    You don't need to find an inverse function to answer this question, you only need to determine the value of its derivative at 2. I am supposing that you know the chain rule:

    Let [itex]y=g(x)[/itex] so that [itex]g^{-1}(y)=x[/itex]. Recall that y is a function of x, so in differentiating w.r.t. x we apply the chain rule to get [itex]\left( g^{-1} \right)^{\prime}(y)y^{\prime}=1[/itex] but [itex]y=g(x)[/itex]
    so put [itex]y^{\prime}=g^{\prime}(x)[/itex] and the equation becomes [itex]\left( g^{-1} \right)^{\prime}\left( g(x)\right) g^{\prime}(x)=1[/itex] or [itex]g^{\prime}(x)=\frac{1}{\left( g^{-1} \right)^{\prime}\left( g(x)\right)}[/itex].

    In your problem, let [itex]g(x)=f^{-1}(x)[/itex].
     
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