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

To find \( \frac{d y}{d x} \) for \( y=\sqrt{x}(x^{2}+3)^{5} \), we will use the product rule and the chain rule. Let \( u = \sqrt{x} \) and \( v = (x^{2}+3)^{5} \). 1. Differentiate \( u \): \[ \frac{du}{dx} = \frac{1}{2\sqrt{x}} \] 2. Differentiate \( v \) using the chain rule: - First, differentiate the outer function: \[ \frac{dv}{dx} = 5(x^{2}+3)^{4} \cdot \frac{d}{dx}(x^{2}+3) \] - Next, differentiate the inner function: \[ \frac{d}{dx}(x^{2}+3) = 2x \] - So: \[ \frac{dv}{dx} = 5(x^{2}+3)^{4} \cdot 2x = 10x(x^{2}+3)^{4} \] Now, apply the product rule: \[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \] Substituting in \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \): \[ \frac{dy}{dx} = \sqrt{x} \cdot 10x(x^{2}+3)^{4} + (x^{2}+3)^{5} \cdot \frac{1}{2\sqrt{x}} \] Now, simplify: 1. First term: \[ \frac{dy}{dx} = 10x\sqrt{x}(x^{2}+3)^{4} \] 2. Second term: \[ \frac{1}{2\sqrt{x}}(x^{2}+3)^{5} \] Final derivative will be: \[ \frac{dy}{dx} = 10x\sqrt{x}(x^{2}+3)^{4} + \frac{1}{2\sqrt{x}}(x^{2}+3)^{5} \] This gives you the derivative in terms of \(x\).