Question

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

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

Answer 1

The correct option is B $\frac{\sec {}^{2}x+1}{2y}$

$y=\sqrt{\tan x+\sqrt{\tan x}}$

$y=\sqrt{\tan x+y}$

$y^2-y=\tan x$

$=02\left(\frac{dy}{dx}\right)y-1={\sec }^{2}x$

$2\left(\frac{dy}{dx}\right)y=\frac{{\sec }^{2}x+1}{2}$

$\frac{dy}{dx}=\frac{{\sec }^{2}x+1}{2y}$