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Thread: Hyperbolic Functions #2

  1. #1
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    Question Hyperbolic Functions #2

    $\displaystyle f(x) = arccosh\frac{1}{\sqrt{x^2-7x+x}} $
    #1. Find the derivative of the function.
    #2. State the interval where the function is differentiable and calculate the derivative.

    here is my attempted solution so far:
    For $\displaystyle y = cosh^{-1}x$, the domain is $\displaystyle x \geq 1. $ So,
    $\displaystyle \frac{1}{\sqrt{x^2-7x+x}} \geq 1 $

    $\displaystyle 1 \geq \sqrt{x^2-7x+x} $

    $\displaystyle 1 \geq x^2 - 6x - 1 $

    $\displaystyle \frac{6-\sqrt{40}}{2} \leq x \leq \frac{6+\sqrt{40}}{2} $

    also $\displaystyle \sqrt{x^2-7x+x} \geq 0 $

    $\displaystyle x^2 - 6x \geq 0 $

    $\displaystyle x \geq 0, x \geq 6 $


    I am also stumped at how to solve the second part of this question. Thanks for any help!
    Last edited by xxlvh; Oct 26th 2009 at 02:38 PM. Reason: LaTex errors
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  2. #2
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    For starters, you have a serious typo. Is the polynomial under the radical $\displaystyle x^2-7x-x$, $\displaystyle x^2-6x-1$, or $\displaystyle x^2-6x$? You interchange among all three in your question.

    Secondly, rewrite $\displaystyle f(x)=a(b(c(x)))$ where $\displaystyle a(x)=\cosh^{-1}(x), b(x)=x^{-1/2}, c(x)=x^2-7x-x$, and you can derive using the chain rule: $\displaystyle f'(x)=a'(b(c(x)))b'(c(x))c'(x)$
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