1
$\begingroup$

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?

  • 0
    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.2012-11-28
  • 0
    @DonAntonio, that's what I was thinking. I can get it using my Ti-89, but to use the definition seems ridiculous.2012-11-28
  • 1
    I suggest you use the definition of derivative to derive all the formulas you need first, then use them...2012-11-28
  • 0
    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.2012-11-28
  • 0
    You can use this [technique](http://math.stackexchange.com/questions/244055/monotonicity-of-function-fx-11-xx/244140#244140) to prove the function is increasing.2012-11-28
  • 0
    Thanks for the comments guys, but I'm still lost as to how to use the definition formula to get the derivative. Still at the point mentioned of the numerator.2012-11-28
  • 0
    `show that f is monotonically increasing`... Not interested anymore?2012-12-01

1 Answers 1

1

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.