Question

If $y = \sqrt{\tan x + \sqrt{\tan x + \sqrt{\tan x + \dots \infty}}}$, then $(2y - 1)\frac{dy}{dx} =$

• A. $\sec^2 x$
• B. $-\sec^2 x$
• C. $\tan x$
• D. $\sec x \tan x$

Hint:

For any infinite nested function of the form $y = \sqrt{f(x) + y}$, the derivative is always $\frac{dy}{dx} = \frac{f'(x)}{2y-1}$.

Answer 1

The Correct Option is A

Step 1: Concept
For an infinite nested radical of the form $y = \sqrt{f(x) + \sqrt{f(x) + \dots \infty}}$, we can write it as $y = \sqrt{f(x) + y}$.

Step 2: Meaning
Squaring both sides of the equation gives us a simplified relation that can be differentiated using implicit differentiation.

Step 3: Analysis
Squaring both sides: \[ y^2 = \tan x + y \] Differentiating both sides with respect to $x$: \[ 2y \frac{dy}{dx} = \sec^2 x + \frac{dy}{dx} \] Rearranging terms to group $\frac{dy}{dx}$: \[ 2y \frac{dy}{dx} - \frac{dy}{dx} = \sec^2 x \] \[ (2y - 1)\frac{dy}{dx} = \sec^2 x \]

Step 4: Conclusion
The value of the expression is $\sec^2 x$.

Final Answer: (A)