Question

Let $y=x^{x^{x.......}},$ then $\frac{dy}{dx}$ is equal to

• A. $y{x}^{y-1}$
• B. $\frac{y^2}{x(1-ylogx)}$
• C. $\frac{y}{x(1+ylogx)}$
• D. None of these

Answer 1

The correct option is B $\frac{y^2}{x(1-ylogx)}$

We have $y=x^{x^{x.....}}$
As y has repeated infinite powers of $x$ so it can be written as $y=x^y$
Taking log of both sides, we get $\log y=y.\log x$
Differentiating with respect to $x$

$\frac{1}{y}.\frac{dy}{dx}=y.\frac{1}{x}+\frac{dy}{dx}.logx$
$\therefore \frac{dy}{dx}=\frac{y}{x}.\frac{y}{(1-ylogx)}=\frac{y^2}{x(1-y.logx)}$