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Thread: limit definition of the derivative problem

  1. #1
    VkL
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    limit definition of the derivative problem

    Use algebra and the limit definition of the derivative to
    find f′(−1), when

    $\displaystyle
    f(x) = \frac {2}{1 -5x}
    $


    I know the limit definition of the derivative is:
    as the Lim h-->0 [F(x+a)-f(x)]/h
    I get confused on the actual agebra part.
    If anyone can kindly work it out, that would be very nice. thank you.
    Last edited by VkL; Oct 16th 2008 at 08:52 PM.
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  2. #2
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    $\displaystyle f\left( x \right) = \frac{2}
    {{1 - 5x}} \hfill \\$

    $\displaystyle f\left( {x + h} \right) = \frac{2}
    {{1 - 5\left( {x + h} \right)}} \hfill \\$

    $\displaystyle f\left( {x + h} \right) - f\left( x \right) = \frac{2}
    {{1 - 5\left( {x + h} \right)}} - \frac{2}
    {{1 - 5x}} \hfill \\$

    $\displaystyle = \frac{2}
    {{1 - 5x - 5h}} - \frac{2}
    {{1 - 5x}} \hfill \\$

    $\displaystyle = \frac{{2\left( {1 - 5x} \right) - 2\left( {1 - 5x - 5h} \right)}}
    {{\left( {1 - 5x - 5h} \right)\left( {1 - 5x} \right)}} \hfill \\$

    $\displaystyle = \frac{{2 - 10x - 2 + 10x + 10h}}
    {{\left( {1 - 5x - 5h} \right)\left( {1 - 5x} \right)}} \hfill \\$

    $\displaystyle = \frac{{10h}}
    {{\left( {1 - 5x - 5h} \right)\left( {1 - 5x} \right)}} \hfill \\$

    $\displaystyle {\text{Now,}} \hfill \\$

    $\displaystyle f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}
    {h} = \mathop {\lim }\limits_{h \to 0} \frac{{10h}}
    {{h\left( {1 - 5x - 5h} \right)\left( {1 - 5x} \right)}} \hfill \\$

    $\displaystyle = \mathop {\lim }\limits_{h \to 0} \frac{{10}}
    {{\left( {1 - 5x - 5h} \right)\left( {1 - 5x} \right)}} = \frac{{10}}
    {{\left( {1 - 5x - 0} \right)\left( {1 - 5x} \right)}} \hfill \\$

    $\displaystyle = \frac{{10}}
    {{\left( {1 - 5x} \right)\left( {1 - 5x} \right)}} = \frac{{10}}
    {{\left( {1 - 5x} \right)^2 }} \hfill \\$

    $\displaystyle f'\left( x \right) = \frac{{10}}
    {{\left( {1 - 5x} \right)^2 }} \hfill \\$

    $\displaystyle f'\left( { - 1} \right) = \frac{{10}}
    {{\left[ {1 - 5\left( { - 1} \right)} \right]^2 }} = \frac{{10}}
    {{36}} = \frac{5}
    {{18}} \hfill \\ $
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  3. #3
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by VkL View Post
    Use algebra and the limit definition of the derivative to
    find f′(−1), when



    f(x) = 2/(1-5x)


    I know the limit definition of the derivative is:
    as the Lim h-->0 [F(x+a)-f(x)]/h
    I get confused on the actual agebra part.
    If anyone can kindly work it out, that would be very nice. thank you.

    $\displaystyle \lim_{h \to 0}~\frac{\frac{2}{1-5(x+h)} - \frac{2}{1-5x}}{h}$


    $\displaystyle \lim_{h \to 0}~\frac{\frac{2}{1-5x-5h} - \frac{2}{1-5x}}{h}$


    $\displaystyle \lim_{h \to 0}~ \frac{\frac{2(1-5x) - 2(1-5x-5h)}{(1-5x-5h)(1-5x)}}{h}$

    $\displaystyle \lim_{h \to 0}~\frac{\frac{2-10x-2+10x+10h}{1-5x-5h-5x+25x^2+25xh}}{h}$


    $\displaystyle \lim_{h \to 0}~ \frac{\frac{10h}{1-10x+25x^2-5h+25xh}}{h}$


    $\displaystyle \lim_{h \to 0}~ \frac{10h}{h(1-10x+25x^2-5h+25xh)}$


    $\displaystyle \lim_{h \to 0}~ \frac{10}{1-10x+25x^2-5h+25xh}$


    $\displaystyle \frac{10}{1-10x+25x^2}$
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  4. #4
    VkL
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    which 1 is correct

    which 1 is correct?
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  5. #5
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    Quote Originally Posted by VkL View Post
    which 1 is correct?
    They are both correct.

    Because $\displaystyle 1-10x+25x^2 = (1-5x)^2$.

    Thus, $\displaystyle f'(x) = \frac{10}{(1-5x)^2}$.

    Now let $\displaystyle x=1$ thus,
    $\displaystyle f'(1) = \frac{10}{(1-5)^2} = \frac{10}{16}$.
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