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Thread: inverse function differentiation

  1. #1
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    inverse function differentiation

    Let
    $\displaystyle F(x) = \int_{1}^{x}{(1+t^2)^{\frac{1}{2}}dt}$ , $\displaystyle 1 \le x$ . Let G be the inverse function of F. Evaluate G'(0).

    I have $\displaystyle F'(x)=(1+x^2)^{\frac{1}{2}}$ for $\displaystyle 1 \le x$ but I am having trouble finding G(x)
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    Quote Originally Posted by 234578 View Post
    Let
    $\displaystyle F(x) = \int_{1}^{x}{(1+t^2)^{\frac{1}{2}}dt}$ , $\displaystyle 1 \le x$ . Let G be the inverse function of F. Evaluate G'(0).

    I have $\displaystyle F'(x)=(1+x^2)^{\frac{1}{2}}$ for $\displaystyle 1 \le x$ but I am having trouble finding G(x)
    Dear 234578,

    Try integrating the expression and obtain F(x) using the substitution, $\displaystyle t=tan\theta $

    Afterwards you could find the inverse of F(x) and G'(0)

    Hope this will help you.
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  3. #3
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    Quote Originally Posted by 234578 View Post
    Let
    $\displaystyle F(x) = \int_{1}^{x}{(1+t^2)^{\frac{1}{2}}dt}$ , $\displaystyle 1 \le x$ . Let G be the inverse function of F. Evaluate G'(0).

    I have $\displaystyle F'(x)=(1+x^2)^{\frac{1}{2}}$ for $\displaystyle 1 \le x$ but I am having trouble finding G(x)
    since F and G are inverses ...

    $\displaystyle F[G(x)] = x$

    $\displaystyle F'[G(x)] \cdot G'(x) = 1$

    $\displaystyle G'(x) = \frac{1}{F'[G(x)]}$

    $\displaystyle G'(0) = \frac{1}{F'[G(0)]}$

    since $\displaystyle F(1) = 0$ , $\displaystyle G(0) = 1$

    $\displaystyle G'(0) = \frac{1}{F'(1)}$
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