Question

If $y=\sqrt{\tan x+\sqrt{\tan x+\sqrt{\tan x+\ldots \infty}}}$ then $\frac{d y}{d x}=$

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

Answer 1

Answer: (B)

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