Explanation
Let $$y = y_1+y_2$$ $$\Rightarrow \cfrac{dy}{dx}=\cfrac{dy_1}{dx}+\cfrac{dy_2}{dx}$$ Now $$y_1 = \left(1+\frac{1}{x}\right)^x \Rightarrow \log y_1 =x \log\left(1+\frac{1}{x}\right)$$
Differentiating both side w.r.t $$x$$
$$\Rightarrow \cfrac{1}{y_1}\cfrac{dy_1}{dx} = \log(1+\frac{1}{x})+x.\cfrac{1}{1+\frac{1}{x}}.\left(\frac{-1}{x}^2\right)$$
$$\Rightarrow \cfrac{dy_1}{dx} = y_1\left[\log(1+\frac{1}{x})-\cfrac{1}{1+x}\right]$$ And $$y_2 = x^{1+\frac{1}{x}} \Rightarrow \log y_2 = (1+\frac{1}{x}) \log x$$
Differentiating both sides,
$$\Rightarrow \cfrac{1}{y_2}\cfrac{dy_2}{dx} = (1+\frac{1}{x})\cfrac{1}{x}+(\log x)(\frac{-1}{x}^2)$$
$$\Rightarrow \cfrac{dy_2}{dx} = y_2\left[\cfrac{(1+x)- \log x}{x^2}\right]$$ $$\therefore \cfrac{dy}{dx} = \left(1+\cfrac{1}{x}\right)\left[\log(1+\frac{1}{x})-\cfrac{1}{1+x}\right]+x^{1+\frac{1}{x}}\left[\cfrac{(1+x)- \log x}{x^2}\right]$$
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