Question

(a) If \( y = \sqrt{\cos x + y} \), prove that \[ \frac{dy}{dx} = \frac{\sin x}{1 - 2y}. \]

Hint:

Implicit differentiation is key for solving equations where \( y \) depends on \( x \) and appears on both sides of the equation.

Answer 1

The provided equation is: \[ y = \sqrt{\cos x + y}. \] Squaring both sides yields: \[ y^2 = \cos x + y. \] Differentiating with respect to \( x \) gives: \[ 2y \frac{dy}{dx} = -\sin x + \frac{dy}{dx}. \] Rearranging to isolate \( \frac{dy}{dx} \)-terms: \[ 2y \frac{dy}{dx} - \frac{dy}{dx} = -\sin x. \] Factoring out \( \frac{dy}{dx} \): \[ \frac{dy}{dx} (2y - 1) = -\sin x. \] Solving for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{-\sin x}{2y - 1}. \] Given that \( y = \sqrt{\cos x + y} \), \( y \) must be non-negative. Thus, the expression can be simplified to: \[ \frac{dy}{dx} = \frac{\sin x}{1 - 2y}. \] Therefore, it is proven that: \[ \frac{dy}{dx} = \frac{\sin x}{1 - 2y}. \]