Question

If $y={(1+\frac{1}{x})}^x$ then $\frac{2\sqrt{y_2\left(2\right)+\frac{1}{8}}}{(\log \frac{3}{2}-\frac{1}{3})}$ equal to

Answer 1

Taking logarithm on both sides, we get
$\log y=x\left[\log (1+\frac{1}{x})\right]$
$\Rightarrow$ $\frac{1}{y}{y}_1\left(x\right)=\frac{x^2}{x+1}(-\frac{1}{x^2})+\log (1+\frac{1}{x})=-\frac{1}{x+1}+\log (1+\frac{1}{x})$ ---------(i)
Since $y\left(2\right)={(1+\frac{1}{2})}^2=\frac{9}{4}$ so
$y_1\left(2\right)=\left(\frac{9}{4}\right)(-\frac{1}{3}+\log \frac{3}{2})$
MUltiplying (i) by y and then differentiating, we get
$y_2\left(x\right)=y_1\left(x\right)(-\frac{1}{x+1}+\log (1+\frac{1}{x}))+y\left(x\right)\times (\frac{1}{{(x+1)}^2}+\frac{x}{x+1}(-\frac{1}{x^2}))$
So, $y_2\left(2\right)=y_1\left(2\right)(-\frac{1}{3}+\log \frac{3}{2})+y\left(2\right)(\frac{1}{9}-\frac{1}{6})$
$=\left(\frac{9}{4}\right){(-\frac{1}{3}+\log \frac{3}{2})}^2-\frac{1}{8}$.
$\Rightarrow y_2\left(2\right)=\left(\frac{9}{4}\right)(-\frac{1}{3}+\log \frac{3}{2}{)}^2-\frac{1}{8}$
$\Rightarrow \frac{2\sqrt{y_2\left(2\right)+\frac{1}{8}}}{\log \frac{3}{2}-\frac{1}{3}}=3$