Results 1 to 5 of 5

Thread: limit definition of the derivative problem

  1. #1
    VkL
    VkL is offline
    Member
    Joined
    Oct 2008
    Posts
    96

    limit definition of the derivative problem

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

    <br />
f(x) = \frac {2}{1 -5x}<br />


    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Aug 2008
    Posts
    530

    Reply

     f\left( x \right) = \frac{2}<br />
{{1 - 5x}} \hfill \\

    f\left( {x + h} \right) = \frac{2}<br />
{{1 - 5\left( {x + h} \right)}} \hfill \\

    f\left( {x + h} \right) - f\left( x \right) = \frac{2}<br />
{{1 - 5\left( {x + h} \right)}} - \frac{2}<br />
{{1 - 5x}} \hfill \\

    = \frac{2}<br />
{{1 - 5x - 5h}} - \frac{2}<br />
{{1 - 5x}} \hfill \\

    = \frac{{2\left( {1 - 5x} \right) - 2\left( {1 - 5x - 5h} \right)}}<br />
{{\left( {1 - 5x - 5h} \right)\left( {1 - 5x} \right)}} \hfill \\

    = \frac{{2 - 10x - 2 + 10x + 10h}}<br />
{{\left( {1 - 5x - 5h} \right)\left( {1 - 5x} \right)}} \hfill \\

    = \frac{{10h}}<br />
{{\left( {1 - 5x - 5h} \right)\left( {1 - 5x} \right)}} \hfill \\

    {\text{Now,}} \hfill \\

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

    = \mathop {\lim }\limits_{h \to 0} \frac{{10}}<br />
{{\left( {1 - 5x - 5h} \right)\left( {1 - 5x} \right)}} = \frac{{10}}<br />
{{\left( {1 - 5x - 0} \right)\left( {1 - 5x} \right)}} \hfill \\

    = \frac{{10}}<br />
{{\left( {1 - 5x} \right)\left( {1 - 5x} \right)}} = \frac{{10}}<br />
{{\left( {1 - 5x} \right)^2 }} \hfill \\

    f'\left( x \right) = \frac{{10}}<br />
{{\left( {1 - 5x} \right)^2 }} \hfill \\

    f'\left( { - 1} \right) = \frac{{10}}<br />
{{\left[ {1 - 5\left( { - 1} \right)} \right]^2 }} = \frac{{10}}<br />
{{36}} = \frac{5}<br />
{{18}} \hfill \\
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member 11rdc11's Avatar
    Joined
    Jul 2007
    From
    New Orleans
    Posts
    894
    Thanks
    1
    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.

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


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


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

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


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


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


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


    \frac{10}{1-10x+25x^2}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    VkL
    VkL is offline
    Member
    Joined
    Oct 2008
    Posts
    96

    which 1 is correct

    which 1 is correct?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by VkL View Post
    which 1 is correct?
    They are both correct.

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

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

    Now let x=1 thus,
    f'(1) = \frac{10}{(1-5)^2} = \frac{10}{16}.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] limit definition of derivative
    Posted in the Calculus Forum
    Replies: 6
    Last Post: Mar 3rd 2011, 01:53 PM
  2. limit definition of a derivative?
    Posted in the Calculus Forum
    Replies: 8
    Last Post: Dec 5th 2009, 09:57 PM
  3. Finding derivative from limit definition
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 29th 2009, 08:40 PM
  4. Limit definition and derivative
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Feb 13th 2009, 01:08 PM
  5. derivative using limit definition
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Oct 5th 2008, 08:18 AM

Search Tags


/mathhelpforum @mathhelpforum