Differentiating a QuotientDate: 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 Check out our web site! http://mathforum.org/dr.math/ |
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