Find \( \frac{d y}{d x} \) 5. \( y=\sqrt{x}\left(x^{2}+3\right)^{5} \)

To find \( \frac{dy}{dx} \) for the function \( y = \sqrt{x}(x^{2} + 3)^{5} \), we'll use the product rule and chain rule. 1. Identify \( u = \sqrt{x} \) and \( v = (x^{2} + 3)^{5} \). 2. Calculate \( \frac{du}{dx} \) and \( \frac{dv}{dx} \): \[ \frac{du}{dx} = \frac{1}{2\sqrt{x}} \] \[ \frac{dv}{dx} = 5(x^{2}+3)^{4} \cdot \frac{d}{dx}(x^{2}+3) = 5(x^{2}+3)^{4} \cdot 2x = 10x(x^{2}+3)^{4} \] 3. Apply the product rule: \[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \] Substituting our values in: \[ \frac{dy}{dx} = \sqrt{x} \cdot 10x(x^{2}+3)^{4} + (x^{2}+3)^{5} \cdot \frac{1}{2\sqrt{x}} \] 4. Simplifying, we can combine the terms: \[ \frac{dy}{dx} = 10x\sqrt{x}(x^{2}+3)^{4} + \frac{(x^{2}+3)^{5}}{2\sqrt{x}} \] And that gives us the derivative \( \frac{dy}{dx} \).