Question

If $y=\sqrt{\tan x+\sqrt{\tan x+\sqrt{\tan x+\dots \infty }}}$ then $\frac{dy}{dx}=$

• A. $\frac{{\cos }^{2}x}{2y-1}$
• B. $\frac{{\sec }^{2}x}{2y-1}$
• C. $\frac{\tan x}{2y-1}$
• D. $\frac{\cot x}{2y-1}$

Answer 1

Answer: (B)

$y=\sqrt{\tan x+y}$
$\Rightarrow y^2=\tan x+y$
$\Rightarrow 2y\cdot \frac{dy}{dx}={\sec }^{2}x+\frac{dy}{dx}$
$\Rightarrow \frac{dy}{dx}=\frac{{\sec }^{2}x}{2y-1}$