Question

If \(y = \sqrt{tanx+\sqrt{tanx+\sqrt{tanx+\cdots \cdots \cdots \infty }}}\), then \((\frac{dy}{dx})^2\) at \(x = \frac{π}{4}\) is

• A. \(\frac{5}{4}\)
• B. \(\frac{-5}{4}\)
• C. \(\frac{4}{5}\)
• D. \(\frac{-4}{5}\)

Hint:

Since the root repeats, $y^2=\tan x+y$.

Answer 1

The Correct Option is C

Step 1: Use the repeating structure
The expression under the first root is \(\tan x+y\) itself, so \(y^2=\tan x+y\).

Step 2: Differentiate
\(2yy'=\sec^2x+y'\), so \(y'=\frac{\sec^2x}{2y-1}\).

Step 3: Find \(y\) at \(x=\frac{\pi}{4}\)
\(\tan\frac{\pi}{4}=1\), so \(y^2-y-1=0\) and \(y=\frac{1+\sqrt5}{2}\) (positive root).
Then \(2y-1=\sqrt5\) and \(\sec^2\frac{\pi}{4}=2\).

Step 4: Square
\(y'=\frac{2}{\sqrt5}\), so \((y')^2=\frac{4}{5}\).

Final Answer:
\((y')^2=\frac45\), option (C). \[ \boxed{\text{(C) } \frac{4}{5}} \]