Question

$\mathrm{If}y=x^{x^{x^{.....\infty }}},\mathrm{then}\frac{\mathrm{dy}}{\mathrm{dx}}\mathrm{at}y=x=1\mathrm{is}$

• A. 1
• B. -1
• C. \N
• D. $\frac{1}{2}$

Answer 1

The correct option is A 1

Given that, $y=x^{x^{x^{.....\infty }}}$
Since by deleting a single term from an infinite series, it remains same. Therefore, the given function may be written as
$y=x^y\Rightarrow \log y=y\log x\Rightarrow \frac{\log y}{y}=\log x\Rightarrow \frac{y\cdot \frac{1}{y}\frac{\mathrm{dy}}{\mathrm{dx}}-\log y\frac{\mathrm{dy}}{\mathrm{dx}}}{y^2}=\frac{1}{x}\Rightarrow \left(1-\log y\right)\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{y^2}{x}\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{y^2}{x\left(1-\log y\right)}\mathrm{At}x=1\mathrm{and}y=1,\Rightarrow {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{y=x=1}=\frac{1}{1\left(1-0\right)}=1$