Question

If \[ y=\sqrt{x+\sqrt{x+\sqrt{x+\cdots\infty}}}, \] then \[ \frac{dy}{dx}= \]

• A. \(\frac{1}{2y}\)
• B. \(\frac{1}{1-2y}\)
• C. \(\frac{1}{2(1-2y)}\)
• D. \(\frac{-1}{1-2y}\)

Hint:

For infinite radicals, replace the repeated radical part by the original variable expression.

Answer 1

The Correct Option is D

Concept: In infinite radical expressions, the repeated part is again equal to \(y\).

Step 1: Given: \[ y=\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}} \] Since the expression under the first square root repeats itself, we can write: \[ y=\sqrt{x+y} \]

Step 2: Square both sides. \[ y^2=x+y \]

Step 3: Differentiate both sides with respect to \(x\). \[ \frac{d}{dx}(y^2)=\frac{d}{dx}(x+y) \] \[ 2y\frac{dy}{dx}=1+\frac{dy}{dx} \]

Step 4: Bring derivative terms together. \[ 2y\frac{dy}{dx}-\frac{dy}{dx}=1 \] \[ (2y-1)\frac{dy}{dx}=1 \]

Step 5: Solve for \(\frac{dy}{dx}\). \[ \frac{dy}{dx}=\frac{1}{2y-1} \] This can also be written as: \[ \frac{dy}{dx}=\frac{-1}{1-2y} \] Therefore, \[ \boxed{\frac{-1}{1-2y}} \]