Challenging differentiation problem

Question

I am trying to do it with substitution and getting a $1$ as answer but the answer is $\ln (4e^2)$

Given: $$ y = (1 + \frac{1}{x})^x + x^{1+\frac{1}{x}}$$

Evaluate $y'(1)$

Answer 1

First , we consider the following :- $$g=(x+\frac{1}{x})^x \implies \ln g = {x}\ln{(x+\frac{1}{x})}\implies \frac{1}{g} \cdot \frac{dg}{dx}=\ln(x+\frac{1}{x})+x\cdot \frac{x}{x^2+1}\cdot(1-\frac{1}{x^2})\implies g’=(x+\frac{1}{x})^x\cdot\left(\ln(x+\frac{1}{x})+x\cdot \frac{x}{x^2+1}\cdot(1-\frac{1}{x^2})\right)$$ $$ \therefore g’(1) = 2 \cdot \ln 2 $$

$$h=x^{1+\frac{1}{x}}\implies \ln h = (1+\frac{1}{x})\cdot \ln x \implies \frac{1}{h}\cdot\frac{dh}{dx} =-\frac{1}{x^2}\cdot\ln x+(1+\frac{1}{x})\cdot\frac{1}{x} \implies h’= x^{1+\frac{1}{x}} \left(-\frac{1}{x^2}\cdot \ln x+(1+\frac{1}{x})\cdot\frac{1}{x} \right)$$ $$ \therefore h’(1)=2$$

From the above relations , we get $$y’(1)=h’(1)+g’(1)=2+2\cdot( \ln2)=\boxed{\ln(4 \cdot e^2)}$$