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John E. Franklin
2006-02-21 12:56:19

You probably see your mistake now, you have typed - 1/h, when you mean - 1/a

So in the overall numerator, there are two fractions subtracted, get like denominators.
Then subtract their numerators.
Then you'll see  (-h/x(x+h))/h and the h's cancel and the limit can be taken.  Good Luck!

John E. Franklin
2006-02-21 12:47:24

Anchor, the very first definition of derivative means delta y divided by delta x, which means rise over run or slope.
limit where delta goes to zero.  You forgot the f(x + h)  and f(x), the f makes it the y part.
So 1/(x+h) - 1/x all divided by h is how you start off.
I'll work on it...

ryos
2006-02-21 08:05:19

Or, using the power rule:

(1/x) = x^-1
(x^-1)′ = -1(x^-2) = -1/x^2

kempos
2006-02-21 03:26:40

[f(x)/g(x)]' = (f'(x)*g(x) - f(x)*g'(x))/[g(x)]^2 = (0*x - 1*1)/x^2 = -1/x^2

anchor
2006-02-20 15:24:36

the problem gives the function of x as: f(x)= 1/x

Im confused as to the steps that lead to the answer: -1/a^2

i know the formula for finding this is :

lim                (a+h)-(a)/h
h->0
i can get to here just fine -> ( 1/(a+h) - (1/h) ) / h

but somehow i always wind up getting -h/(a^2+ah)
what am i doing wrong?

any help is appreciated,