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Math Help - Finding Derivative of a Quotient involving Radicals

  1. #1
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    Finding Derivative of a Quotient involving Radicals

    Alright so the problem simply says "Find the derivative of each of the following functions" and one listed is as follows:

    q(x)=((3x-5)^(.5))/((x+4)^(.5))

    it is written q(x)= √3x-5 / √x+4

    [I'm on a Mac, didn't want to use only symbols in case they show up weird for PC]
    Last edited by Eternal Calculus Student; November 29th 2009 at 04:39 PM. Reason: typo
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Eternal Calculus Student View Post
    Alright so the problem simply says "Find the derivative of each of the following functions" and one listed is as follows:

    q(x)=((3x-5)^(.5))/((x+4)^(.5))

    it is written q(x)= √3x-5 / √x+4

    [I'm on a Mac, didn't want to use only symbols in case they show up weird for PC]
    q\!\left(x\right)=\sqrt{\frac{3x-5}{x+4}}.

    I would recommnd using quotient rule in conjunction with chain rule. I'll get you started:

    \begin{aligned}q^{\prime}\!\left(x\right)&=\frac{1  }{2}\left(\frac{3x-5}{x+4}\right)^{-1/2}\cdot\frac{\,d}{\,dx}\left[\frac{3x-5}{x+4}\right]\\ &= \frac{1}{2}\sqrt{\frac{x+4}{3x-5}}\cdot\frac{\,d}{\,dx}\left[\frac{3x-5}{x+4}\right]\\ &= \ldots\end{aligned}

    Now apply quotient rule and simplify.

    Can you take it from here?
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  3. #3
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    Oh, definitely, I didn't realize that it would be that simple the other way around. I confused myself doing the quotient rule first @.@ I should be fine now thank you so much
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  4. #4
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    Quote Originally Posted by Eternal Calculus Student View Post
    Alright so the problem simply says "Find the derivative of each of the following functions" and one listed is as follows:

    q(x)=((3x-5)^(.5))/((x+4)^(.5))

    it is written q(x)= √3x-5 / √x+4

    [I'm on a Mac, didn't want to use only symbols in case they show up weird for PC]
    q(x) = \left(\frac{3x-5}{x+4}\right)^{\frac{1}{2}}

    q'(x) = \frac{1}{2} \left(\frac{3x-5}{x+4}\right)^{-\frac{1}{2}} \cdot \frac{3(x+4) - (3x-5)}{(x+4)^2}

    you can clean up the algebra ...
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