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Math Help - Find the derivative

  1. #1
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    Find the derivative

    Darn professors....wish mine explained things...

    How do I find f`(a) for  f(x) = (3x +1)^{1/2}

    Thats the square root of 3x+1, double thanks to someone who shows me how to put that in there.

    I have only seen f(x) - f(a)/x-a and f(a + h) - f(a)/h, so no dx or whatever that stuff is please.
    Last edited by mr fantastic; February 5th 2009 at 10:04 PM. Reason: Fixed the latex
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  2. #2
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    Think what can you multiply the expression by that will lead to the expression to a difference of two squares?
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  3. #3
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    Quote Originally Posted by Oiram View Post
    Think what can you multiply the expression by that will lead to the expression to a difference of two squares?
    what?! assume im brain dead cause i basically am.
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  4. #4
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    Check this out:
    Difference of two squares - Wikipedia, the free encyclopedia
    Although the expression may be under the square root operation you may still use the technique found in the link above.
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  5. #5
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    Quote Originally Posted by TastyBeverage View Post
    Darn professors....wish mine explained things...

    How do I find f`(a) for  f(x) = (3x +1)^{1/2}

    Thats the square root of 3x+1, double thanks to someone who shows me how to put that in there.

    I have only seen f(x) - f(a)/x-a and f(a + h) - f(a)/h, so no dx or whatever that stuff is please.
    \lim_{x \rightarrow a} \frac{\sqrt{3x + 1} - \sqrt{3a + 1}}{x - a} = \lim_{x \rightarrow a} \left( \frac{\sqrt{3x + 1} - \sqrt{3a + 1}}{x - a}\right) \cdot \left( \frac{ \sqrt{3x + 1} + \sqrt{3a + 1} }{\sqrt{3x + 1} + \sqrt{3a + 1}} \right) = \, ....
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  6. #6
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    Quote Originally Posted by mr fantastic View Post
    \lim_{x \rightarrow a} \frac{\sqrt{3x + 1} - \sqrt{3a + 1}}{x - a} = \lim_{x \rightarrow a} \left( \frac{\sqrt{3x + 1} - \sqrt{3a + 1}}{x - a}\right) \cdot \left( \frac{ \sqrt{3x + 1} + \sqrt{3a + 1} }{\sqrt{3x + 1} + \sqrt{3a + 1}} \right) = \, ....
    ahhh....rationalize. Thanks!
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