How do I find the derivative of $(1 +1/x)^x $ I tried one approach but the correction in the book shows me a total different answer.
Here's what I did:
$(1+ 1/x)^x=xln(1+1/x)$
Thus, now we try to find the derivative of a multiplication:
$ u(x)=x$ 
$(u(x))'=1$
$v(x)=ln(1+1/x)$ 
$(v(x))'= -1/(x^2) +1/x$
And so:
$(uv)'=u'v+uv'$
which gives us:
$(uv)'=xln(1 +1/x) +(-1/x^2 -1/x)$
Yet, the correction gives me this as an answer:
$ln(1+1/x)-1/(1+x)$

Let $f(x) = (1+1/x)^x$. Then $$\ln(f(x)) = x \ln(1 + 1/x) = x \ln( (x+1)/x) = x \ln(x+1) - x \ln(x).$$
Taking derivatives on both sides gives:
$$\frac{f'(x)}{f(x)} = \ln(x+1) + \frac{x}{x+1} - \ln(x) - 1.$$
We wish to find $f'(x)$, so we solve for it and subsitute the value of $f(x)$.
$$f'(x) = \left(1 + \frac{1}{x} \right)^x \left[ \ln(x+1) + \frac{x}{x+1} - \ln(x) - 1 \right]=\left(1 + \frac{1}{x} \right)^x \left[ \ln\left( 1 + \frac{1}{x} \right) - \frac{1}{x+1} \right]$$
