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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

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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.

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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.

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  • $\begingroup$ Also, be sure and mind your negative sign in the numerator when you find that common denominator and combine your expression. This is a point where many people make a mistake. $\endgroup$ – Zeta10 Jan 7 '15 at 15:46
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Good job so far. Using your idea you get

$$\lim_{\Delta x \to 0} \frac{1}{\Delta x} \frac{x+1 - [x + \Delta x + 1]}{((x + \Delta x) + 1)(x+1)} = \lim_{\Delta x \to 0} \frac{ - 1}{((x + \Delta x) + 1)(x+1)} = -\frac{1}{(x+1)^2}$$

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