Question

Differentiate $y={\sqrt{x}}^{{\sqrt{x}}^{...\infty }}$

• A. $\frac{dy}{dx}=\frac{y^3}{x(2-y\log x)}$
• B. $\frac{dy}{dx}=\frac{y^2}{x(2-y\log x)}$
• C. $\frac{dy}{dx}=\frac{y}{x(2-y\log x)}$
• D. None of these

Answer 1

The correct option is B $\frac{dy}{dx}=\frac{y^2}{x(2-y\log x)}$

$y={\sqrt{x}}^{{\sqrt{x}}^{...\infty }}$

$y={\sqrt{x}}^y$

$\ln y=y\ln \sqrt{x}$

Differentiating wrt $x$

$\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{y}{\sqrt{x}}\frac{1}{2}\left(x^{\frac{-1}{2}}\right)+\frac{dy}{dx}\ln \sqrt{x}$

$\Rightarrow \frac{y^{\prime }}{y}-y^{\prime }\ln \sqrt{x}=\frac{y}{2x}$

$y^{\prime }(1-y\ln \sqrt{x})=\frac{y^2}{2x}$

$y^{\prime }=\frac{y^2}{x(2-y\ln x)}$