Results 1 to 2 of 2

Math Help - Finding a derivative

  1. #1
    Junior Member
    Joined
    Sep 2009
    Posts
    62

    Finding a derivative

    The problem is this:

    Find the derivative of the function f(x) = \frac{-2}{(x + 1)^2} using a difference quotient. Hint: leave the denominator in factored form.

    Also, I should mention, the only way my class has learned to solve for this so far is using the equation: f'(x) = \lim h \rightarrow 0 \frac{f(x + h) - f(x)}{h}

    So, I believe I've solved the problem, but the way in which I did it seems incorrect, even though it yields the correct answer (or maybe there's a better way...I just feel like my way isn't what would be preferred!)

    Here's what I did:

    1.) f'(x) = \lim h \rightarrow 0 \frac{f(x + h) - f(x)}{h}

    2.) f'(x) = \lim h \rightarrow 0 \frac{ \frac{-2}{(x + h + 1)^2} - \frac{-2}{(x + 1)^2}  }{h}

    3.) I then found a common denominator and combined the top. I also made the h on the bottom into \frac {1}{h} to make it a little easier to write. That gave me this:

    f'(x) = \lim h \rightarrow 0   (\frac{1}{h})    <br />
(\frac{2(x + h + 1)^2 - (2(x + 1)^2)}{(x + 1)^2 (x + h + 1)^2})<br />

    4.) For the sake of not writing out all the algebra, I multiplied the entire top part through and ended up with this:

    f'(x) = \lim h \rightarrow 0 (\frac{1}{h})  (\frac{2h^2 + 4hx + 4h}{(x + 1)^2(x + h + 1)^2})<br />

    5.) Then I factored out an "h" from the top and canceled it with the 1/h, getting this:

    f'(x) = \lim h \rightarrow 0 (\frac{2h + 4x + 4}{(x + 1)^2(x + h + 1)^2})<br />

    6.) Now here's where I think what I've done might not be considered correct. I got stuck once I reached this point, so I attempted to just take the limit of all of that at once like this (sorry for how long this is!):

    f'(x) =\frac{ <br />
\lim h \rightarrow 0  (2h) + \lim h \rightarrow 0  (4x) + \lim h \rightarrow 0  (4)}<br />
{[\lim h \rightarrow 0  (x + 1)]^2[\lim h \rightarrow 0 (x) + \lim h \rightarrow 0  (h) + \lim h \rightarrow 0  (1)]^2}<br />


    Which, after figuring all of that out gave me this:
    <br />
\frac{2(0) + 4x + 4}{(x+1)^2(x + 0 + 1)(x + 0 + 1)}

    If you simplify that you get

    \frac{4x + 4}{(x + 1)^2 (x + 1)^2}

    Factor out a 4 from the top, cancel, and you get a final answer of

    \frac {4}{(x + 1)^3}


    Now, that's the right answer but...am I getting to it correctly? That's my question.

    Any help at all would be much appreciated! I know it's a really long problem.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Amer's Avatar
    Joined
    May 2009
    From
    Jordan
    Posts
    1,093
    Quote Originally Posted by lysserloo View Post
    The problem is this:

    Find the derivative of the function f(x) = \frac{-2}{(x + 1)^2} using a difference quotient. Hint: leave the denominator in factored form.

    Also, I should mention, the only way my class has learned to solve for this so far is using the equation: f'(x) = \lim h \rightarrow 0 \frac{f(x + h) - f(x)}{h}

    So, I believe I've solved the problem, but the way in which I did it seems incorrect, even though it yields the correct answer (or maybe there's a better way...I just feel like my way isn't what would be preferred!)

    Here's what I did:

    1.) f'(x) = \lim h \rightarrow 0 \frac{f(x + h) - f(x)}{h}

    2.) f'(x) = \lim h \rightarrow 0 \frac{ \frac{-2}{(x + h + 1)^2} - \frac{-2}{(x + 1)^2}  }{h}

    3.) I then found a common denominator and combined the top. I also made the h on the bottom into \frac {1}{h} to make it a little easier to write. That gave me this:

    f'(x) = \lim h \rightarrow 0   (\frac{1}{h})    <br />
(\frac{2(x + h + 1)^2 - (2(x + 1)^2)}{(x + 1)^2 (x + h + 1)^2})<br />

    4.) For the sake of not writing out all the algebra, I multiplied the entire top part through and ended up with this:

    f'(x) = \lim h \rightarrow 0 (\frac{1}{h})  (\frac{2h^2 + 4hx + 4h}{(x + 1)^2(x + h + 1)^2})<br />

    5.) Then I factored out an "h" from the top and canceled it with the 1/h, getting this:

    f'(x) = \lim h \rightarrow 0 (\frac{2h + 4x + 4}{(x + 1)^2(x + h + 1)^2})<br />

    6.) Now here's where I think what I've done might not be considered correct. I got stuck once I reached this point, so I attempted to just take the limit of all of that at once like this (sorry for how long this is!):

    f'(x) =\frac{ <br />
\lim h \rightarrow 0  (2h) + \lim h \rightarrow 0  (4x) + \lim h \rightarrow 0  (4)}<br />
{[\lim h \rightarrow 0  (x + 1)]^2[\lim h \rightarrow 0 (x) + \lim h \rightarrow 0  (h) + \lim h \rightarrow 0  (1)]^2}<br />


    Which, after figuring all of that out gave me this:
    <br />
\frac{2(0) + 4x + 4}{(x+1)^2(x + 0 + 1)(x + 0 + 1)}

    If you simplify that you get

    \frac{4x + 4}{(x + 1)^2 (x + 1)^2}

    Factor out a 4 from the top, cancel, and you get a final answer of

    \frac {4}{(x + 1)^3}


    Now, that's the right answer but...am I getting to it correctly? That's my question.

    Any help at all would be much appreciated! I know it's a really long problem.

    you solution is correct but in the 6th step there is no need to separate the limit to all just sub h=0 just

    and write the limit like this \lim_{h\rightarrow 0} to be like this \lim_{h\rightarrow 0} or like this lim_{h\rightarrow 0} to be like lim_{h\rightarrow 0}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Help finding derivative
    Posted in the Calculus Forum
    Replies: 6
    Last Post: June 29th 2010, 08:00 PM
  2. Finding the Derivative...
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 2nd 2010, 08:39 AM
  3. Finding the third derivative
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 18th 2009, 07:43 PM
  4. Finding the derivative
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 3rd 2009, 07:57 AM
  5. Finding the value of the derivative.
    Posted in the Calculus Forum
    Replies: 6
    Last Post: February 10th 2008, 10:17 AM

Search Tags


/mathhelpforum @mathhelpforum