\( y=f(\{u\}) \) find dexivutive \( \frac{d y}{d x} \) \( y=\sqrt{x+\sqrt{x+\sqrt{x}}} \)

To find the derivative \( \frac{dy}{dx} \) for the function \( y = \sqrt{x + \sqrt{x + \sqrt{x}}} \), we can utilize the chain rule and the power rule. To begin, let's define the innermost function and work our way outward. Let \( z = \sqrt{x} \), then substituting we have: 1. \( y = \sqrt{x + \sqrt{x + \sqrt{x}}} \) 2. \( = \sqrt{x + \sqrt{x + z}} \) Now we can differentiate: - First, we differentiate \( \sqrt{x + \sqrt{x + \sqrt{x}}} \) using \( \frac{d}{dx} (x^{1/2}) = \frac{1}{2} x^{-1/2} \). - Then, we apply the chain rule for each layer of the nested square roots. We can systematically derive: 1. Inner Function: \( u = x + \sqrt{x + \sqrt{x}} \) 2. Find \( \frac{du}{dx} \). 3. Outer Function: \( y = \sqrt{u} \) gives \( \frac{dy}{du} = \frac{1}{2\sqrt{u}} \). Combine using the chain rule: \[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \] When executed correctly, this will lead to the derivative \( \frac{dy}{dx} \) through substitution of \( u \). Remember to simplify your final answer accordingly to avoid any common mistakes like skipping the steps in differentiating nested functions!