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 will need to use the product rule and the chain rule. Let \( u = \sqrt{x} \) and \( v = (x^2 + 3)^5 \). Then \( y = u \cdot v \). First, find the derivatives of \( u \) and \( v \): 1. \( u = x^{1/2} \), so \( \frac{du}{dx} = \frac{1}{2} x^{-1/2} \). 2. For \( v = (x^2 + 3)^5 \), we apply the chain rule: \( \frac{dv}{dx} = 5(x^2 + 3)^4 \cdot (2x) = 10x(x^2 + 3)^4 \). Now, apply the product rule: \[ \frac{dy}{dx} = u \cdot \frac{dv}{dx} + v \cdot \frac{du}{dx} \] Substituting the derivatives we found and the expressions for \( u \) and \( v \): \[ \frac{dy}{dx} = \sqrt{x} \cdot 10x(x^2 + 3)^4 + (x^2 + 3)^5 \cdot \frac{1}{2} x^{-1/2} \] This can be simplified further. Thus, the derivative \( \frac{dy}{dx} \) is: \[ \frac{dy}{dx} = 10x^{3/2}(x^2 + 3)^4 + \frac{1}{2\sqrt{x}}(x^2 + 3)^5 \]