# Finding Derivatives $f(x)={1\over x+1}$

I'm using the Limit Definition to find the derivative,

$$f'(x)=\lim_{\Delta x \to 0} {f(x+\Delta x) - f(x) \over \Delta x}$$  Now, I want to find the derivative for the function,

$$f(x)={1 \over x+1}$$

So, here's what I did.

$$\lim_{\Delta x \to 0} {{1 \over (x+\Delta x) +1} - {1 \over x+1}\over \Delta x}$$

Now, I think I can multiply the numerator and the denominator by the least common multiple to get rid of the denominator in the numerator??

I'm not sure what to do from here. Thanks

Once you've done what you've said you think you can do (which is to expand the large fraction by $(x + 1)(x + \Delta x + 1)$, the least common multiple of the small denominators), you get $$\frac{(x+1) - (x + \Delta x +1)}{\Delta x(x+1)(x+\Delta x + 1)} = \frac{-\Delta x}{\Delta x(x+1)(x+\Delta x + 1)} = -\frac{1}{(x+1)(x+\Delta x + 1)}$$ Now take the limit as $\Delta x \to 0$, and you get your result.
You are correct. You can actually factor out the $\frac{1}{\triangle x}$ that is in the denominator of this complex rational expression and then do all of your simplification.
$$\lim_{\Delta x \to 0} \frac{1}{\Delta x} \frac{x+1 - [x + \Delta x + 1]}{((x + \Delta x) + 1)(x+1)} = \lim_{\Delta x \to 0} \frac{ - 1}{((x + \Delta x) + 1)(x+1)} = -\frac{1}{(x+1)^2}$$