${\left( {x + \cfrac{1}{x}} \right)^x} + {x^{\left( {1 + \cfrac{1}{x}} \right)}}$

Let $y = {\left( {x + \cfrac{1}{x}} \right)^x} + {x^{\left( {1 + \cfrac{1}{x}} \right)}} = u + v$
where $u = {\left[ {x + \cfrac{1}{x}} \right]^x}$ and $v = {x^{\left( {1 + \cfrac{1}{x}} \right)}}$

$\Rightarrow$ $\cfrac{{dy}}{{dx}} = \cfrac{{du}}{{dx}} + \cfrac{{dv}}{{dx}}$ ….(i)

Now, $u = {\left( {x + \cfrac{1}{x}} \right)^x} \Rightarrow \log u = x\log \left( {x + \cfrac{1}{x}} \right)$ …..(ii)

Differentiating (ii) w.r.t. x, we get

$\Rightarrow$ $\cfrac{1}{u}\cfrac{{du}}{{dx}} = x\cfrac{d}{{dx}}\log \left( {x + \cfrac{1}{x}} \right) + \log \left( {x + \cfrac{1}{x}} \right)(1)$

$= \cfrac{x}{{x + \cfrac{1}{x}}}\left( {1 - \cfrac{1}{{{x^2}}}} \right) + \log \left( {x + \cfrac{1}{x}} \right)$

$\Rightarrow$ $\cfrac{{du}}{{dx}} = {\left( {x + \cfrac{1}{x}} \right)^x}\left[ {\cfrac{x}{{x + \cfrac{1}{x}}}\left( {x - \cfrac{1}{{{x^2}}}} \right) + \log \left( {x + \cfrac{1}{x}} \right)} \right]$ ….(iii)

Also, $v = {x^{\left( {x + \cfrac{1}{x}} \right)}}$

Taking log on both the sides, we get

$\Rightarrow$ $\log v = \left( {x + \cfrac{1}{x}} \right)\log x$ …(iv)

Differentiating (iv) on both sides w.r.t. x, we get

$\Rightarrow$ $\cfrac{1}{v}\cfrac{{dv}}{{dx}} = \left( {x + \cfrac{1}{x}} \right)\cfrac{d}{{dx}}\log x + \log x\cfrac{d}{{dx}}\left( {x + \cfrac{1}{x}} \right)$

$= \left( {x + \cfrac{1}{x}} \right)\cfrac{1}{x} + \log x\left( { - \cfrac{1}{{{x^2}}}} \right)$

$\Rightarrow$ $\cfrac{{dv}}{{dx}} = {x^{\left( {x + \cfrac{1}{x}} \right)}}\left[ {\left( {x + \cfrac{1}{x}} \right)\cfrac{1}{x} + \log x\left( {\cfrac{{ - 1}}{{{x^2}}}} \right)} \right]$ …(v)

Substituting the values of (iii) and (v) in (i), we get

$\cfrac{{dy}}{{dx}} = {\left( {x + \cfrac{1}{x}} \right)^x}\left[ {\cfrac{x}{{x + \cfrac{1}{x}}}\left( {x - \cfrac{1}{{{x^2}}}} \right) + \log \left( {x + \cfrac{1}{x}} \right)} \right]$ $+ {x^{\left( {x + \cfrac{1}{x}} \right)}}\left[ {\left( {x + \cfrac{1}{x}} \right)\cfrac{1}{x} + \log x\left( {\cfrac{{ - 1}}{{{x^2}}}} \right)} \right]$

$= {\left( {x + \cfrac{1}{x}} \right)^x}\left[ {\cfrac{{{x^2} - 1}}{{{x^2} + 1}} + \log \left( {x + \cfrac{1}{x}} \right)} \right] + {x^{\left( {x + \cfrac{1}{x}} \right)}}\left[ {\cfrac{{x + 1 - \log x}}{{{x^2}}}} \right]$