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**anchor****Member**- Registered: 2006-02-19
- Posts: 1

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,

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**kempos****Member**- Registered: 2006-01-07
- Posts: 77

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

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**ryos****Member**- Registered: 2005-08-04
- Posts: 394

Or, using the power rule:

(1/x) = x^-1

(x^-1)′ = -1(x^-2) = -1/x^2

El que pega primero pega dos veces.

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,561

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...

**igloo** **myrtilles** **fourmis**

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,561

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!

**igloo** **myrtilles** **fourmis**

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