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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 $y'(1)$

Sinπ
  • 830
Akshat
  • 9

1 Answers1

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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 $$y’(1)=h’(1)+g’(1)=2+2\cdot( \ln2)=\boxed{\ln(4 \cdot e^2)}$$

Sinπ
  • 830
  • For $h'\left(1\right)$, note that $h\left(1\right) = 1^2 = 1$. The right side using the derivative gives $\ln x = 0$, so the $1$ term is $0$ and you just need to use $\left(1 + \frac{1}{x}\right) \cdot \frac{1}{x}$. However, with $x = 1$, it gives $2$, not $1$. As such, I get an answer of $2 + 2 \cdot \left(\ln 2\right) = \ln\left(4 \cdot e^2\right)$ instead. – John Omielan Jan 31 '19 at 07:54
  • @JohnOmielan Yes you are right . Ty for spotting the error. Will fix it – Sinπ Jan 31 '19 at 10:31