Question:hard

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

Show Hint

For any infinite nested function of the form $y = \sqrt{f(x) + y}$, the derivative is always $\frac{dy}{dx} = \frac{f'(x)}{2y-1}$.
Updated On: Jun 3, 2026
  • $\sec^2 x$
  • $-\sec^2 x$
  • $\tan x$
  • $\sec x \tan x$
Show Solution

The Correct Option is A

Solution and Explanation

Step 1: Use the self-repeating trick.
The expression under the first root repeats forever: it is $\tan x$ plus the same whole thing again. So the entire expression equals $y$. That lets us write $y = \sqrt{\tan x + y}$.

Step 2: Square both sides.
Squaring removes the outer root and gives a clean equation.
\[ y^2 = \tan x + y \]

Step 3: Get ready to differentiate.
Now differentiate both sides with respect to $x$. We treat $y$ as a function of $x$, so its derivative is $\frac{dy}{dx}$.

Step 4: Differentiate each term.
The left $y^2$ gives $2y\frac{dy}{dx}$. The $\tan x$ gives $\sec^2 x$, and the $y$ gives $\frac{dy}{dx}$.
\[ 2y\frac{dy}{dx} = \sec^2 x + \frac{dy}{dx} \]

Step 5: Collect the $\frac{dy}{dx}$ terms.
Move the $\frac{dy}{dx}$ from the right to the left.
\[ 2y\frac{dy}{dx} - \frac{dy}{dx} = \sec^2 x \]

Step 6: Factor out $\frac{dy}{dx}$.
Take the common factor on the left.
\[ (2y - 1)\frac{dy}{dx} = \sec^2 x \]
\[ \boxed{\sec^2 x} \]
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