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Thread: Inverse Derivative Help

  1. #1
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    Question Inverse Derivative Help

    I cannot figure out what to do!!
    Given the function $\displaystyle f(x)=5 x^3+2 x+5$ Let g be the inverse function of f. i.e. $\displaystyle g(x)=f^{-1}(x).$
    $\displaystyle g^{\prime}(12)=$

    I can't manage to find g(x) let alone g'(x)
    Thank you for any help.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Krooger View Post
    I cannot figure out what to do!!
    Given the function $\displaystyle f(x)=5 x^3+2 x+5$ Let g be the inverse function of f. i.e. $\displaystyle g(x)=f^{-1}(x).$
    $\displaystyle g^{\prime}(12)=$

    I can't manage to find g(x) let alone g'(x)
    Thank you for any help.
    note that $\displaystyle g'(x) = \frac 1{f'(g(x))}$ .....not hard to derive. your textbook should do this

    now, by definition, $\displaystyle g(a) = x \implies f(x) = a$

    thus, $\displaystyle g(12)$ is the x-value so that $\displaystyle f(x) = 12$

    by inspection, this is 1, thus $\displaystyle g(12) = 1$


    thus, $\displaystyle g'(12) = \frac 1{f'(1)}$

    i leave the rest to you
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    note that $\displaystyle g'(x) = \frac 1{f'(g(x))}$ .....not hard to derive. your textbook should do this

    now, by definition, $\displaystyle g(a) = x \implies f(x) = a$

    thus, $\displaystyle g(12)$ is the x-value so that $\displaystyle f(x) = 12$

    by inspection, this is 1, thus $\displaystyle g(12) = 1$


    thus, $\displaystyle g'(12) = \frac 1{f'(1)}$

    i leave the rest to you
    $\displaystyle f^{-1}(x) \neq \frac{1}{f(x)}$.

    Edit: Never mind, I figured out what you were doing. I worked it out using implicit differentiation, but your way works too.
    Last edited by Prove It; Jan 25th 2009 at 02:52 AM.
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  4. #4
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    Quote Originally Posted by Prove It View Post
    $\displaystyle f^{-1}(x) \neq \frac{1}{f(x)}$.

    Edit: Never mind, I figured out what you were doing. I worked it out using implicit differentiation, but your way works too.
    The first formula Jhevon gives comes from implicit differentiation of $\displaystyle f(g(x))=x$

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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Prove It View Post
    $\displaystyle f^{-1}(x) \neq \frac{1}{f(x)}$.
    that's not the formula i have

    Edit: Never mind, I figured out what you were doing. I worked it out using implicit differentiation, but your way works too.
    yes, i worked it out that way as well...er, like Moo said
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  6. #6
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    I was unaware that:
    $\displaystyle
    g'(x) = \frac 1{f'(g(x))}$

    ...and to be honest upon first glace it still dosen't make sense to me haha. When I get a minute I will have to look it up. Thank You all for the help.
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Krooger View Post
    I was unaware that:
    $\displaystyle
    g'(x) = \frac 1{f'(g(x))}$

    ...and to be honest upon first glace it still dosen't make sense to me haha. When I get a minute I will have to look it up. Thank You all for the help.
    since $\displaystyle f(x)$ and $\displaystyle g(x)$ are inverse functions,

    $\displaystyle f(g(x)) = x$

    now differentiate both sides with respect to x (note you will need the chain rule to differentiate the left side) and solve for $\displaystyle g'(x)$
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