Question

If \( y = \sqrt{\tan{x} + \sqrt{\tan{x} + \sqrt{\tan{x} + \dots}}} \), then show that \[ \frac{dy}{dx} = \sec^2{x} \cdot \frac{2y - 1}{y^2 - 1} \] Find \( \frac{dy}{dx} \) at \( x = 0 \).

Hint:

When dealing with nested square roots, set up an equation and differentiate implicitly.

Answer 1

Step 1: Let \( y = \sqrt{\tan{x} + \sqrt{\tan{x} + \dots}} \). This implies: \[ y = \sqrt{\tan{x} + y} \] Step 2: Square both sides: \[ y^2 = \tan{x} + y \] Step 3: Rearrange the equation: \[ y^2 - y - \tan{x} = 0 \] Step 4: Differentiate implicitly with respect to \( x \): \[ 2y \frac{dy}{dx} - \frac{dy}{dx} = \sec^2{x} \] Step 5: Factor out \( \frac{dy}{dx} \): \[ \frac{dy}{dx} (2y - 1) = \sec^2{x} \] Step 6: Solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{\sec^2{x}}{2y - 1} \] Step 7: Substitute \( y^2 - 1 = (y - 1)(y + 1) \): \[ \frac{dy}{dx} = \sec^2{x} \cdot \frac{2y - 1}{y^2 - 1} \] Step 8: Evaluate at \( x = 0 \): At \( x = 0 \), \( \tan{0} = 0 \), so \( y = 0 \). Thus: \[ \frac{dy}{dx} = \frac{1}{2 \cdot 0 - 1} = -1 \] Thus, \( \frac{dy}{dx} \) at \( x = 0 \) is: \[ \boxed{-1} \]