Question

lf $y^{y^{y^{{.}^{.\infty }}}}=lo{g}_e(x+lo{g}_e(x+\dots \left)\right)$ then $\frac{dy}{dx}$ at$(x=e^2-2,y=\sqrt{2})$ is

• A. $\frac{\log \left(e/2\right)}{2\sqrt{2}(e^2-1)}$
• B. $\frac{\log 2}{2\sqrt{2}(e^2-1)}$
• C. $\frac{\sqrt{2}\log \left(e/2\right)}{(e^2-1)}$
• D. $\frac{\log \left(e/2\right)}{(e^2-1)}$

Answer 1

The correct option is A $\frac{\log \left(e/2\right)}{2\sqrt{2}(e^2-1)}$

Let $y^{y^{y^{{.}^{.....\infty }}}}=lo{g}_e(x+lo{g}_e(x+.....\infty \left)\right)=u$

Then , $u=y^u$ and $u={\log }_e(x+u)$

Taking, $u=y^u$

$\log u=u\log y$

$\frac{1}{u}=u\left(\frac{1}{y}\right)\frac{dy}{du}+\log y$

$\frac{1}{u}-\log y=\frac{u}{y}\frac{dy}{du}$

$\frac{y}{u^2}-\frac{y}{u}\log y=\frac{dy}{du}$

$\Rightarrow \frac{y}{u^2}(1-u\log y)=\frac{dy}{du}$

$\Rightarrow u^{\left(1/u\right)-2}(1-\log u)=\frac{dy}{du}$.....(1)

Now, taking $u={\log }_e(x+u)$

$1=\frac{1}{(x+u)}(\frac{dx}{du}+1)$

$\frac{dx}{du}=x+u-1$

$\Rightarrow \frac{dx}{du}=e^u-1$...(2)

Since $y=u^{1/u}$, when $y=\sqrt{2}$, $u=2$ or $u=4$

Also since $x=e^u-u$, when $x=e^2-2$, $u=2$.

The value of $u$ which satisfies both is $u=2$

From (1) and (2) $\frac{dy}{dx}$ at the given values of 'x' and 'y' is

$\frac{2^{-3/2}(1-\log 2)}{e^2-1}$

$=\frac{\log \left(e/2\right)}{2\sqrt{2}(e^2-1)}$