Differentiating $f(x)= \sqrt[x]{x}$ from first principles An interesting fact is that the point of maximum of $y=\sqrt[x]{x}$ is attained when x equals to exactly e (as seen from the pic)
I know how to differentiate $y= \sqrt[x]{x}$, by logarithmic methodology, but can anyone teach me how to differentiate this function using only the first principle?
Note: I am still a beginner, thus it would help greatly if you can include all the steps.

 A: I'd consider the case $x>0$. Since the functions $\exp:{\Bbb R}\rightarrow {\Bbb R}_{>0}: x\mapsto e^x$ and $\ln:{\Bbb R}_{>0}\rightarrow {\Bbb R}: x\mapsto \ln(x)$ are inverse to each other, we have $\sqrt[x]x = x^{1/x} = e^{\ln(x^{1/x})}$. From here you should be able to differentiate w.r.t. $x$.
A: A nice trick is to rewrite $y=\sqrt[x]x$ as $$y=e^{\frac{\ln(x)}{x}}\tag{1}$$ 
This is because $$y=\sqrt[x]x=x^\frac{1}{x}=e^{\ln\left(x^\frac{1}{x}\right)}=e^\frac{\ln(x)}{x}$$
Differentiating $(1)$, and setting it equal to $0$ yields: $$\frac{dy}{dx}=e^{\frac{\ln(x)}{x}}\left(\frac{1}{x^2}-\frac{\ln(x)}{x^2}\right)=0\\\Rightarrow \frac{1}{x^2}-\frac{\ln(x)}{x^2}=0\\\Rightarrow \ln(x)=1\\\Rightarrow x=e$$
A: You have the "outer" function $$\Psi(u,v):=u^v\qquad(u\in{\mathbb R}_{>0}, \ v\in{\mathbb R})$$ of two variables, with partial derivatives
$$\Psi_u(u,v)=v\,u^{v-1}\>,\qquad\Psi_v(u,v)=\log (u)\ u^v\ .$$
The function $$f(x):=\Psi\left(x,{1\over x}\right)$$ then has to be differentiated using the chain rule:
$$\eqalign{f'(x)&=\Psi_u(x,1/x)\cdot 1+\Psi_v(x,1/x)\cdot\left(-{1\over x^2}\right)\cr  &={1\over x}x^{1/x-1}+\log (x)\>x^{1/x}\left(-{1\over x^2}\right)\cr   &={\root  x \of x\over x^2}\bigl(1-\log (x)\bigr)\ .\cr}$$
