Question

If $y=\sqrt{\tan x+\sqrt{\tan x+\sqrt{\tan x+..\mathrm{to}\infty }}}$, prove that $\frac{dy}{dx}=\frac{{\sec }^{2}x}{2y-1}$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have},y=\sqrt{\tan x+\sqrt{\tan x+\sqrt{\tan x+...\mathrm{to}\infty }}} \\ \Rightarrow y=\sqrt{\tan x+y} \\ \mathrm{Squaring}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}, \\ {y}^2=\tan x+y \\ \Rightarrow 2y\frac{dy}{dx}={\sec }^{2}x+\frac{dy}{dx} \\ \Rightarrow \frac{dy}{dx}\left(2y-1\right)={\sec }^{2}x \\ \Rightarrow \frac{dy}{dx}=\frac{{\sec }^{2}x}{2y-1} \end{gathered}$$