I am trying to show that the function $f(x)=(1+\frac1x)^x$ on $[0,+\infty)$.
I have found that $f'(x)=f(x) \left[\ln\left(\frac{x+1}x\right)- \frac1{x+1}\right]$.
Since $f(x)$ is always positive, I only have to show that $$\ln\left(\frac{x+1}x\right)>\frac1{x+1}\text{ when }x>0.$$
Is there an easy way to do this?
A different approach altogether:
If we start with Bernoulli's inequality, $(1+u)^r\gt1+ru$ for $u\ge0$ and $r\ge1$ (which is easy to prove by taking the derivative of $f(u)=(1+u)^r-1-ru$), we have, on letting $u=rx$,
$$\left(1+{1\over xr}\right)^r\ge1+r\cdot{1\over rx}=1+{1\over x}\implies\left(1+{1\over rx}\right)^{rx}=\left(\left(1+{1\over rx}\right)^r\right)^x\ge\left(1+{1\over x}\right)^x\quad\text{if }r\ge1$$
hence if $y=rx\ge x$ then
$$\left(1+{1\over y}\right)^y\ge\left(1+{1\over x}\right)^x$$
$$ \begin{align} \log\left(1+\frac1x\right) &=\int_0^{1/x}\frac{\mathrm{d}t}{1+t}\\ &\ge\int_0^{1/x}\frac{\mathrm{d}t}{1+1/x}\\[3pt] &=\frac1{x+1} \end{align} $$
$$\ln\frac{x+1}x>\frac1{x+1}$$ $$\iff-\ln\left(1-\frac1{x+1}\right)>\frac1{x+1}$$ For $x>0$, $0<\frac1{x+1}<1$, so substitute $y=\frac1{x+1}$: $$\iff-\ln(1-y)>y$$ The Maclaurin series of $\ln(1-y)$ is always valid for $0<y<1$: $$\iff y+\frac{y^2}2+\frac{y^2}3+\dots>y$$ $$\iff\frac{y^2}2+\frac{y^3}3+\dots>0$$ which is true since $y$ is positive.
Let $g(x) = \ln\left(\frac{x+1}x\right)-\frac1{x+1} = \ln(x+1)-\ln(x)-\frac1{x+1} $.
$\begin{array}\\ g'(x) &=\frac1{x+1}-\frac1{x}+\frac1{(x+1)^2}\\ &=\frac{x(x+1)-(x+1)^2+x}{x(x+1)^2}\\ &=\frac{-1}{x(x+1)^2}\\ \end{array} $
Therefore $g(x)$ is decreasing.
But $\lim_{x \to \infty} \ln(1+1/x) = 0 $ and $\lim_{x \to \infty} \frac1{1+x} = 0 $, so that $\lim_{x \to \infty} g(x) = 0 $.
Therefore $g(x)> 0$ for $x > 0$.
This is an old problem and there are already clever solutions (see robjhon's solution above using integration) and the comment by Paramanand Sing (using classical inequalities). Both strategies solve the problem under the assumption $x>0$, stated by the OP.
Here I just want to add that the statement is also valid for $x<-1$. The method follows the along the lines proposed by the OP, namely, to look at the sign of $f'(x)$. The whole method relies on a simple extension the inequalities mentioned by Paramanand Sing to all $x+1>0$.
Proposition: If $t+1>0$, then $$\frac{t}{1+t}\leq \log(1+t)\leq t$$ with equality iff $t=0$.
Proof: The right hand side follows immediately as $1+t\leq e^t$ for all $t\in\mathbb{R}$ with equality iff $t=0$. The left hand side can be obtained from $$\frac{1}{1+t}=1-\frac{t}{1+t}\leq \exp\big(-\frac{t}{1+t}\big)$$ by taking logarithms on both sides of this inequality.
Using the computation of $f'(x)=f(x)\Big(\log\big(1+\tfrac{1}{x}\big)-\frac{1}{1+x}\Big)$ already provided by the OP, we obtain that for $x<-1$ or $x>0$, $1+\frac{1}{x}>0$ and so, $$\log\big(1+\tfrac{1}{x}\big)-\frac{1}{1+x}=\log\big(1+\tfrac{1}{x}\big)-\frac{\tfrac1x}{1+\tfrac1x}>0$$
Putting things together, we obtain that $f$ is monotone increasing in $(-\infty,-1)$, and also monotone increasing in $(0,\infty)$.
Comment: The same technique works for showing that $g:x\mapsto \big(1+\frac{1}{x}\big)^{x+1}$ is monotone decreasing: for all $x<-1$ or $x>0$ $$g'(x)=g(x)\Big( \log(1+\tfrac1x)-\tfrac{1}{x}\Big)<0$$ which follows by the proposition above.
All these yield following useful inequality $$\Big(1+\frac{1}{x}\Big)^x<e<\Big(1+\frac{1}{x}\Big)^{x+1}$$ for all $x>0$ and $$\Big(1+\frac{1}{x}\Big)^{x+1}<e<\Big(1+\frac{1}{x}\Big)^x$$ for all $x<-1$.
$$x>0\implies\log (1+1/x)=\log (1+x)-\log x=$$ $$=\int_x^{x+1}(1/t)dt>\int_x^{x+1}(1/(x+1)dt=1/(x+1).$$
Theorem 1. For $x>0$, the function $f_{\alpha}(x)=\bigl(1+\frac1x\bigr)^{x+\alpha}$ increases if and only if $\alpha\le0$ and decreases if and only if $\alpha\ge\frac12$.
For $x<-1$, the function $f_{\alpha}(x)$ decreases if and only if $\alpha\ge1$ and increases if and only if $\alpha\le\frac{1}{2}$.
The necessary and sufficient conditions such that the sequence $a_n=\bigl(1+\frac1n\bigr)^{n+\alpha}$ decreases or increases are $\alpha\ge\frac12$ or $\alpha\le\frac{2\ln3-3\ln2}{2\ln2-\ln3}$ respectively.
Proof. Direct calculation gives \begin{align*} [\ln f_\alpha(x)]'&=\ln\Bigl(1+\frac1x\Bigr) -\frac{x+\alpha}{x(x+1)}& \text{and}& & [\ln f_\alpha(x)]''&=\frac{(2\alpha-1)x+\alpha}{x^2(x+1)^2}. \end{align*}
For $x>0$, it is easy to see that $[\ln f_\alpha(x)]''>0$ and $[\ln f_\alpha(x)]'$ increases if and only if $\alpha\ge\frac12$. Since $\lim_{x\to\infty}[\ln f_\alpha(x)]'=0$ for any $\alpha\in\mathbb{R}$, thus $[\ln f_\alpha(x)]'<0$ for $\alpha\ge\frac12$, which means $f_\alpha'(x)<0$ and $f_\alpha(x)$ decreases. This implies also that the sequence $a_n$ is decreasing for $\alpha\ge\frac12$.
For $x>0$, it is clear that $[\ln f_\alpha(x)]''<0$ and $[\ln f_\alpha(x)]'$ decreases if and only if $\alpha\le0$. Then $[\ln f_\alpha(x)]'>0$, $f_\alpha'(x)>0$ and $f_\alpha(x)$ increases for $\alpha\le0$. This implies that the sequence $\bigl(1+\frac1n\bigr)^{n+\alpha}$ is increasing for $\alpha\le0$.
For $x>0$, when $0<\alpha<\frac12$, the function $[\ln f_\alpha(x)]''$ has a unique zero point $x_0=\frac{\alpha}{1-2\alpha}>0$ which is a supremum point of $[\ln f_\alpha(x)]'$, this supremum equals $[\ln f_{\alpha}(x_0)]'=\ln\bigl(\frac1\alpha-1\bigr)+2(2\alpha-1)>0$. Since $\lim_{x\to0^+}[\ln f_\alpha(x)]'=-\infty$ for $\alpha>0$ and $\lim_{x\to\infty}[\ln f_\alpha(x)]'=0$ for any $\alpha\in\mathbb{R}$, it is yielded that the functions $[\ln f_\alpha(x)]'$ and $f_\alpha'(x)$ have only one zero point $x_1>0$, which is a unique infimum point of $f_\alpha(x)$ on $(0,\infty)$. Consequently, the sufficient and necessary condition of the sequence $a_n$ being increasing is $f_\alpha(1)\le f_\alpha(2)$ which is equivalent to $\alpha\le\frac{2\ln3-3\ln2}{2\ln2-\ln3}$.
For $x<-1$, the function $[\ln f_\alpha(x)]''>0$ and $[\ln f_\alpha(x)]'$ is increasing if and only if $\alpha\le\frac12$. From $\lim_{x\to-\infty}[\ln f_\alpha(x)]'=0$ it is deduced that $[\ln f_\alpha(x)]'>0$ and $f_\alpha'(x)>0$ in $(-\infty,-1)$. Consequently, the function $f_\alpha(x)$ is increasing in $(-\infty,-1)$ if $\alpha\le\frac12$.
For $x<-1$, the function $[\ln f_\alpha(x)]''<0$ and $[\ln f_\alpha(x)]'$ is decreasing if and only if $\alpha\ge1$. From $\lim_{x\to-\infty}[\ln f_\alpha(x)]'=0$ it follows that $[\ln f_\alpha(x)]'<0$ and $f_\alpha'(x)<0$ in $(-\infty,-1)$. Accordingly, the function $f_\alpha(x)$ decreases in $(-\infty,-1)$ if $\alpha\ge1$.
For $x<-1$ and $\frac12<\alpha<1$, the function $[\ln f_\alpha(x)]''$ has a unique zero point $x_0=\frac{\alpha}{1-2\alpha}<-1$ which is a minimum point of $[\ln f_\alpha(x)]'$. Since $\lim_{x\to(-1)^-}[\ln f_\alpha(x)]'=\infty$ and $\lim_{x\to-\infty}[\ln f_\alpha(x)]'=0$, then the functions $[\ln f_\alpha(x)]'$ and $f_\alpha'(x)$ have only one zero point $x_1>0$, which is a unique infimum point of $f_\alpha(x)$ on $(0,\infty)$. This completes the proof of Theorem 1.
Note. The above theorem and its proof are extracted from the paper [3]. For more information on further developments, please refer to the papers [1,2,4] and closely related references therein.
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