# Using definition of derivative to calculate, stuck, cant find a cancellation

1. Jul 27, 2011

### irebat

use the definition of the derivative to compute the derivative of the function at x=1

f(x) = 1/(1 + x 2) for all x.

3. The attempt at a solution
i put it into lim[ (f(x)-f(xo))/x-xo ] form and got

[(1/1+x2) - (1/2)] <---------numerator
___________________________ <----- division sign ("over")
x-1 <----------- denomenator

form and from there cancelled to
2-1-x2 / 2(x2+1)

and then again to -x2 + 1 / 2(x2+1)

but now im stuck and cant find a cancellation to calculate the limit at x=1 without getting a 0 in there soemwhere

if i stick this into the limit calculation formula itll just give me a 0 and thats no good. i already know the answer is -1/2 but i cant quite get there. any help is appreciated, thanks.

2. Jul 27, 2011

### BrianMath

Okay, if I've read this correctly, so far you have
$$\lim_{x\to 1} \frac{\frac{-x^2 + 1}{2(x^2 + 1)}}{x-1}$$
right?

What happens if you factor out a $-1$ from the top fraction's numerator? Is there any algebra you can do on that to divide out the $x-1$?

3. Jul 27, 2011

### irebat

well, actually right now i have

$$\lim_{x\to 1} \frac{-x^2 + 1}{2(x^2 + 1)}$$

4. Jul 27, 2011

### BrianMath

Where did the $x-1$ in the denominator go?

$$\lim_{x\to 1} \frac{\frac{1}{1+x^2} - \frac{1}{2}}{x-1} = \lim_{x\to 1} \frac{\frac{2}{2(1+x^2)} - \frac{1+x^2}{2(1+x^2)}}{x-1} = \lim_{x\to 1} \frac{\frac{2 - 1 - x^2}{2(1+x^2)}}{x-1} = \lim_{x\to 1} \frac{\frac{-x^2 + 1}{2(1+x^2)}}{x-1}$$

5. Jul 27, 2011

### irebat

omg your right.

careless mistake got me again. dang
yeah, taking that -1 out brings out an x-1 in the numerator that cancels with the denominator that I completely forgot about

and then i can take my limit to get -2/4, which is right.

BIG THANKS, man, seriously.

6. Jul 27, 2011

### BrianMath

No problem, little mistakes like these happen to the best of us.

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