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Math Help - Trouble with Derivatives

  1. #1
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    Trouble with Derivatives

    I've learned the short cut to obtain a derivate (nX^n-1) but some questions on exams ask to find the answer using the definition

    Lim f(x+h)-f(x)
    h->0 h

    now I am terribly lost how to solve this is say f(x) = sqrt (x+2)

    what I would think to do is rewrite it was

    (x+2)^1/2 + h) - (x+2)^1/2

    h

    this is where I go astray...
    I am totally baffled on what the next step would be...if it was x^2 I would simply factor it out and then divide by h but the 1/2 is totally making my thought process come to a dead halt. Please tell me if my approach is wrong or what the next step would be. Thanks
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  2. #2
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    Quote Originally Posted by Jay23456 View Post
    I've learned the short cut to obtain a derivate (nX^n-1) but some questions on exams ask to find the answer using the definition

    Lim f(x+h)-f(x)
    h->0 h

    now I am terribly lost how to solve this is say f(x) = sqrt (x+2)

    what I would think to do is rewrite it was

    (x+2)^1/2 + h) - (x+2)^1/2
    h

    this is where I go astray...
    I am totally baffled on what the next step would be...if it was x^2 I would simply factor it out and then divide by h but the 1/2 is totally making my thought process come to a dead halt. Please tell me if my approach is wrong or what the next step would be. Thanks
    f(x) = \sqrt{x + 2}

    f(x + h) = \sqrt{x + h + 2}.



    \frac{f(x + h) - f(x)}{h} = \frac{\sqrt{x + h + 2} - \sqrt{x + 2}}{h}

     = \left(\frac{\sqrt{x + h + 2} - \sqrt{x + 2}}{h}\right)\left( \frac{\sqrt{x + h + 2}+\sqrt{x + 2}}{\sqrt{x + h + 2} + \sqrt{x + 2}}\right)

     = \frac{(x + h + 2) - (x + 2)}{h(\sqrt{x + h + 2} + \sqrt{x + 2})}

     = \frac{h}{h(\sqrt{x + h + 2} + \sqrt{x + 2})}

     = \frac{1}{\sqrt{x + h + 2} + \sqrt{x + 2}}.



    So f'(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}

     = \lim_{h \to 0}\frac{1}{\sqrt{x + h + 2} + \sqrt{x + 2}}

     = \frac{1}{\sqrt{x + 2} + \sqrt{x + 2}}

     = \frac{1}{2\sqrt{x + 2}}.
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  3. #3
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    so when there a square root in the numerator you multiply by the conjugate?
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    Quote Originally Posted by Jay23456 View Post
    so when there a square root in the numerator you multiply by the conjugate?
    Multiply top and bottom by the top's conjugate, yes.
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  5. #5
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    Okay thanks and what would you do if say f(x)= 5/(2-x) ?
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    \lim_{h\to 0}\frac{\frac{5}{2-(x+h)}-\frac{5}{2-x}}{h}

    First simplify \frac{5}{2-(x+h)}- \frac{5}{2- x} by subtracting the fractions- get common denominator by multiplying numerator and denominator of the first fraction by 2- x and the second fraction by 2-x-h:
    \frac{5(2-x)}{(2-x)(2-x-h)}- \frac{5(2-x-h)}{(2-x)(2-x-h)} = \frac{10- 5x- 10+ 5x- 5h}{(2- x)(2- x- h)}.

    Can you finish it?
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  7. #7
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    Where did -5h come from? is -5 times -h = +5h?
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  8. #8
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    Quote Originally Posted by Jay23456 View Post
    Where did -5h come from? is -5 times -h = +5h?
    Yes, it should be +5h.
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  9. #9
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    So I got 5h/(2-x)(2-x-h)

    is that correct? also i'm not sure if the above common denominator is correct shouldn't the denominator be multiplied by (2-x+h)?
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