Question about proving differentiability I'm asked to prove $f\left(x\right)=\left(x\cdot\ln x\right)^{\ln x}$ is differentiable for every x > 1
and calculate it's deriviate.
I have no idea how to approach this question.. Do I need to define a function g that will be the inverse of f?
i.e. $g\left(x\right)=\left(\frac{e^{x}}{x}\right)^{e^{x}}$, prove it's continuous, differential and that it's derivate is not 0 and then use the chain rule?
My attempt:
$f\left(x\right)=\left(x\cdot\ln x\right)^{\ln x}=e^{\ln\left(x\cdot\ln x\right)^{\ln x}}=e^{\left(\ln x\cdot\ln\left(x\cdot\ln x\right)\right)}$
Note that by the algebra of differential functions $e^x,ln(x)$ are differential for every $x > 1$. Therefore $g(x)={\left(\ln x\cdot\ln\left(x\cdot\ln x\right)\right)}$ is also differential for every $x > 1$, and $f$ as a whole is differential for every $x > 1$. Therefore by the chain rule we have
$f'\left(x\right)=\left(e^{\left(\ln x\cdot\ln\left(x\cdot\ln x\right)\right)}\right)^{'}=e^{\left(\ln x\cdot\ln\left(x\cdot\ln x\right)\right)}\cdot\left(\left(\frac{1}{x}\cdot\ln\left(x\cdot\ln x\right)\right)+\left(\frac{1}{x\cdot\ln x}\cdot\left(\ln x+1\right)\right)\cdot\ln x\right)$
$=e^{\left(\ln x\cdot\ln\left(x\cdot\ln x\right)\right)}\cdot\left(\left(\frac{\ln\left(x\cdot\ln x\right)}{x}\right)+\left(\frac{\ln^{2}x+\ln x}{x\cdot\ln x}\right)\right)=e^{\left(\ln x\cdot\ln\left(x\cdot\ln x\right)\right)}\cdot\left(\left(\frac{\ln\left(x\right)+\ln\left(\ln x\right)}{x}\right)+\left(\frac{\ln x+1}{x}\right)\right)$
$=e^{\left(\ln x\cdot\ln\left(x\cdot\ln x\right)\right)}\cdot\left(\left(\frac{\ln\left(x\right)+\ln\left(\ln x\right)+\ln x+1}{x}\right)\right)=e^{\left(\ln x\cdot\ln\left(x\cdot\ln x\right)\right)}\cdot\left(\left(\frac{2\ln\left(x\right)+\ln\left(\ln x\right)+1}{x}\right)\right)=e^{\left(\ln x\cdot\ln\left(x\cdot\ln x\right)\right)}\cdot\left(\left(\frac{\ln\left(x^{2}\right)+\ln\left(\ln x\right)+1}{x}\right)\right)=e^{\left(\ln x\cdot\ln\left(x\cdot\ln x\right)\right)}\cdot\left(\left(\frac{\ln\left(x^{2}\cdot\ln x\right)+1}{x}\right)\right)$
 A: if $x \ge e$ then $\ln x \ge 1$ so $x\cdot \ln  x > 0$ and $\ln (x\cdot \ln  x)$ exists and $x\cdot \ln x = e^{\ln (x\cdot \ln  x)}$.
And then $f(x) = (x\cdot \ln x)^{\ln x}= [e^{\ln (x\cdot \ln  x)}]^{\ln x} = e^{\ln x\dot \ln(x\cdot \ln x)}$.
And that can be differentiated by the using the chain rule and the product rule many times.
If we use the notation $exp(x)= e^x$ and $exp'(x) = $ derivative of $exp(x) =$ derivative of $e^x = e^x=exp(x)$ then
First use of chain rule:  $[exp(\ln x\dot \ln(x\cdot \ln x))]' = exp'(\ln x\dot \ln(x\cdot \ln x))\times (\ln x\dot \ln(x\cdot \ln x))' =$
$exp(\ln x\dot \ln(x\cdot \ln x))\times (\ln x\dot \ln(x\cdot \ln x))'=$
$(x\cdot \ln x)^{\ln x} \times (\ln x\dot \ln(x\cdot \ln x))'=$
Now apply the product rule;
$(x\cdot \ln x)^{\ln x} \times [\ln' x\cdot \ln(x\cdot \ln x) + \ln x (\ln(x\cdot \ln x))'=$
$(x\cdot \ln x)^{\ln x} \times [\frac 1x\cdot \ln(x\cdot \ln x) + \ln x (\ln(x\cdot \ln x))'=$
Now apply the chain rule again:
$(x\cdot \ln x)^{\ln x} \times [\frac 1x\cdot \ln(x\cdot \ln x) + \ln x (\ln'(x\cdot \ln x)\times (x\cdot \ln x)')=$
$(x\cdot \ln x)^{\ln x} \times [\frac 1x\cdot \ln(x\cdot \ln x) + \ln x (\frac 1{x\cdot \ln x}\times (x\cdot \ln x)')=$
And use the product rule again
$(x\cdot \ln x)^{\ln x} \times [\frac 1x\cdot \ln(x\cdot \ln x) + \ln x (\frac 1{x\cdot \ln x}\times (x\cdot (\ln x)'+ (x)'\ln x))=$
$(x\cdot \ln x)^{\ln x} \times [\frac 1x\cdot \ln(x\cdot \ln x) + \ln x (\frac 1{x\cdot \ln x}\times (x\cdot\frac 1x+ 1\cdot\ln x))=$
$(x\cdot \ln x)^{\ln x} \times [\frac 1x\cdot \ln(x\cdot \ln x) + \ln x (\frac 1{x\cdot \ln x}\times (1+ \ln x))=$
$(x\cdot \ln x)^{\ln x}[\frac {\ln(x\cdot\ln x)}x + \ln x(\frac 1{x\ln x} + \frac 1x)]$.
Sof $1 < x < e$ we have $x\ln x < 0$ so $\ln (-x\cdot \ln x)$ exists and $x\cdot \ln x= e^{-\ln(-x\ln x)}$.
And do what we did above.
