Results 1 to 4 of 4

Math Help - Derivative

  1. #1
    Newbie
    Joined
    Jul 2007
    Posts
    19

    Exclamation Derivative

    Hey, guys. Can someone please assist me with...

    In each case, find the derivative dy/dx.
    b.) y=x+1/x-1

    I obtained the answer using the quotient rule, but we are required to use f'(x)= f(x+h)-f(x)/h (which I'm having trouble using for this question).
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by madman1611 View Post
    Hey, guys. Can someone please assist me with...

    In each case, find the derivative dy/dx.
    b.) y=x+1/x-1

    I obtained the answer using the quotient rule, but we are required to use f'(x)= f(x+h)-f(x)/h (which I'm having trouble using for this question).
    This one was actually kind of straight forward

    f(x) = \frac {x + 1}{x - 1}

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

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

    = \lim_{h \to 0} \frac {(x - 1)(x + h + 1) - (x + 1)(x + h - 1)}{h(x + h - 1)(x - 1)}

    = \lim_{h \to 0} \frac {x^2 + xh + x - x - h - 1 - x^2 - xh + x - x - h + 1}{h(x + h - 1)(x - 1)}

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

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

    Now take the limit as h \to 0
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,683
    Thanks
    615
    Hello, madman!

    Find the derivative: . y\:=\:\frac{x+1}{x-1 }
    Defintion: . f'(x) \;=\;\lim_{h\to0}\frac{f(x+h)-f(x)}{h}


    This can be done in four steps: .**

    . . [1] Find f(x+h) . . . replace x with x+h ... and simplify

    . . [2] Subtract f(x) . . . subtract the original function ... and simplify

    . . [3] Divide by h . . . and simplify

    . . [4] Take the limit as h\to0

    . . . . ** .
    Am I the only one that teaches it this way?


    Here we go . . .


    [1]\;\;f(x+h) \;=\;\frac{(x+h) +1}{(x+h)-1}


    [2]\;\;f(x+h)-f(x) \;=\;\frac{x+h+1}{x+h-1} \:- \:\frac{x+1}{x-1}

    . . . . . . . . . . . . . =\;\frac{x-1}{x-1}\cdot\frac{x+h+1}{x+h-1} \:-\:\frac{x+h-1}{x+h-1}\cdot\frac{x+1}{x+1}

    . . . . . . . . . . . . . =\;\frac{(x^2 + xh + x - x - h -  1) - (x^2 + x + xh + h - x - 1)}{(x-1)(x+h-1)}

    . . . . . . . . . . . . . = \;\frac{x^2 + xh + x - x - h - 1 - x^2 - x - xh - h + x + 1}{(x-1)(x+h-1)}

    . . . . . . . . . . . . . =\;\frac{\text{-}2h}{(x-1)(x+h-1)}


    [3]\;\;\frac{f(x+h)-f(x)}{h}\;=\;\frac{\text{-}2h}{h(x-1)(x+h-1)} \;=\;\frac{\text{-}2}{(x-1)(x+h-1)}


    [4]\;\;\lim_{h\to0}\frac{f(x+h)-f(x)}{h} \;=\;\lim_{h\to0}\frac{\text{-}2}{(x-1)(x+h-1)} \;=\;\frac{\text{-}2}{(x-1)(x+0-1)}


    Therefore: . f'(x) \;=\;\frac{\text{-}2}{(x-1)^2}

    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Soroban View Post

    . . . . ** .
    Am I the only one that teaches it this way?
    Apparently so. When I first did calculus (not in the US) no one had trouble with huge formulas, so evaluating it in pieces is not something any teacher thought of doing. In the US, they might have taught it, but usually (in my experience) professors can't pace themselves here, and eventually get into such a rush that your head starts spinning by the time you leave class and you wonder what just happened.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. contuous weak derivative $\Rightarrow$ classic derivative ?
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: April 22nd 2011, 02:37 AM
  2. Replies: 0
    Last Post: January 24th 2011, 11:40 AM
  3. Replies: 2
    Last Post: November 6th 2009, 02:51 PM
  4. Replies: 1
    Last Post: January 7th 2009, 01:59 PM
  5. Fréchet derivative and Gâteaux derivative
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 23rd 2008, 04:40 PM

Search Tags


/mathhelpforum @mathhelpforum