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I have $f(x)=(1+\frac{1}{x})^{x}$, I need to find the derivative of this, using the definition of derivative, and show that f is monotonically increasing.

Using the definition, I have that $$\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h} = \lim_{h\rightarrow0}\frac{(1+\frac{1}{x+h})^{x+h}-(1+\frac{1}{x})^{x}}{h}$$ whose numerator is the same as $(1+\frac{1}{x+h})^{x}(1+\frac{1}{x+h})^{h}-(1+\frac{1}{x})^{x} $

At this point, I'm not sure how to proceed. Would I need to use binomial formula?

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To ask somebody to calculate this function's derivative using the definition seems to me specially evil and even a little satanic. Perhaps there's some nice trick to apply here, but right now I can't see anything making this horror a little less frightening. –  DonAntonio Nov 28 '12 at 4:48
    
@DonAntonio, that's what I was thinking. I can get it using my Ti-89, but to use the definition seems ridiculous. –  Edgar Aroutiounian Nov 28 '12 at 4:50
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I suggest you use the definition of derivative to derive all the formulas you need first, then use them... –  Tunococ Nov 28 '12 at 5:02
    
Well @Dixie, I wouldn't expect a student to calculate this, or any other, derivative with a calculator. In fact, I forbid such things. This derivative though can be achieved in a reasonably easy and fast way using the basic arithmetic of derivatives, the chain rule and etc. –  DonAntonio Nov 28 '12 at 5:04
    
You can use this technique to prove the function is increasing. –  Mhenni Benghorbal Nov 28 '12 at 6:15

1 Answer 1

up vote 1 down vote accepted

For this kind of mess, it is nice to first know the correct answer. The logarithmic derivative seems indicated.

$\ln f(x) = x \ln(1+1/x) = x \ln((x+1)/x) = x \ln(x+1) - x \ln(x)$, so $f'(x)/f(x) = x/(x+1) + \ln(x+1) - 1 - \ln(x) = -1/(x+1) + \ln(1+1/x) $. This tells you that you should get two terms in your result and one of them should involve $\ln(1+1/x)$.

As to how you should continue to get this, I am not sure.

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