Results 1 to 5 of 5

Math Help - derivatives

  1. #1
    Junior Member
    Joined
    Mar 2006
    Posts
    40

    derivatives

    I always have difficulties with fractions...

    Find the derivative of f(x) = (x/x+1) - 3

    I usually break it down to f(x) = x/x +x/1 -3

    f'(x) = 1 + 1

    I know that's not right...please help.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by becky
    I always have difficulties with fractions...

    Find the derivative of f(x) = (x/x+1) - 3
    .
    You can forget about -3 because it is a constant.

    Now,
    \frac{x}{x+1}=\frac{x+1-1}{x+1}=1-\frac{1}{x+1}=1-(x+1)^{-1}
    Thus, using the generalized power rule,
    (x+1)^{-2}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,689
    Thanks
    617
    Hello, becky!

    Find the derivative of: f(x) \:= \:\frac{x}{x+1} - 3

    I usually break it down to: f(x) \:= \:\frac{x}{x} + \frac{x}{1}  -3 . . . I hope not!

    What's wrong with using the Quotient Rule?

    f'(x)\;=\;\frac{(x+1)\cdot1 - x\cdot 1}{(x+1)^2} - 0 \;= \;\frac{1}{(x+1)^2}

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Jun 2006
    From
    San Diego
    Posts
    101
    Hi Becky

    You can only break up a fraction if the added terms are on top, not bottom.

    So while \frac{x+1}{x}=\frac{x}{x}+\frac{1}{x}, it's not true that \frac{x}{x+1} = \frac{x}{x}+\frac{x}{1}.
    So we can't use that kind of techniue in this particular situation.

    What you want to do here is apply the quotient rule, which basically says that if g and h are differentiable functions, then \left(\frac{g}{h}\right)' = \frac{g'h-h'g}{h^2}.

    In your case, g is the top of the fraction x and h is the bottom, x+1.

    So g'=1 and h'=1.

    Then f'(x) = \frac{g'h-h'g}{h^2}-0

    =\frac{1\cdot(x+1)-1\cdot(x)}{(x+1)^2}

    =\frac{1}{(x+1)^2}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Mar 2006
    Posts
    40

    Thanks for your help

    You all explain it so much better than a textbook.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Derivatives and Anti-Derivatives
    Posted in the Calculus Forum
    Replies: 7
    Last Post: February 6th 2011, 06:21 AM
  2. Replies: 1
    Last Post: July 19th 2010, 04:09 PM
  3. Derivatives with both a and y
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 4th 2009, 09:17 AM
  4. Replies: 4
    Last Post: February 10th 2009, 09:54 PM
  5. Trig derivatives/anti-derivatives
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 10th 2009, 01:34 PM

Search Tags


/mathhelpforum @mathhelpforum