Differentiating a Quotient

Date: 08/29/97 at 00:39:06
From: KATHY GLEGHORN
Subject: Functions

Given that g(x) = x-4/x+3, find: {g (x+h) - g(x}/h)

Date: 08/29/97 at 08:33:16
From: Doctor Anthony
Subject: Re: Functions

This is really the way to prove the rule for differentiating a 
quotient - i.e. by letting h -> 0 you will find g'(x).

                    (x+h)-4     x - 4
   g(x+h) - g(x) = --------- - --------
                    (x+h)+3     x + 3

                    (x+3)[x+h-4] - (x-4)(x+h+3)
                 =  ----------------------------
                         (x+3)(x+h+3)


                     (x+3)(x-4) + h(x+3) - (x-4)(x+3) - h(x-4) 
                 =  -------------------------------------------
                          (x+3)(x+h+3)

                      hx + 3h - hx + 4h
                 =   -------------------
                         (x+3)(x+h+3)

                           7h
                 =    ---------------
                       (x+3)(x+h+3)

And dividing by h we get

                            7
                       -------------
                        (x+3)(x+h+3)

If we let h -> 0 then we get

                            7
                        ---------
                         (x+3)^2

Check this result by finding g'(x), using the usual quotient formula

            (x+3) - (x-4)         7
    g'(x) = -------------  =   -------
                (x+3)^2        (x+3)^2


-Doctor Anthony,  The Math Forum
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