\( \begin{array}{l} y=x^{2}(1+x)^{\frac{1}{2}} \\ u=\left(x^{2}\right), \quad v=(1+x)^{\frac{1}{2}} \\ \left.\frac{d u}{d i c}=2 x \quad=\frac{(1+x)^{2}}{d x c} x \quad t-\frac{1}{2}<1+x\right)^{-\frac{1}{2}} \\ \frac{1}{2}(1+x)^{-\frac{1}{2}} \\ u \frac{d v}{d x}+\frac{v d u}{d x} \\ \left(x^{2}\right)_{3} x+(1+x)^{\frac{1}{2}} 2 x \\ x^{3}+\frac{2 x}{\sqrt{1+} x} \end{array} \)

It looks like you're working through the differentiation of the function: \[ y = x^{2}(1+x)^{\frac{1}{2}} \] Let's walk through the differentiation process step-by-step using the **Product Rule**. The Product Rule states that if you have a function \( y = u \cdot v \), then its derivative is: \[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \] ### Step 1: Identify \( u \) and \( v \) Let’s assign: \[ \begin{align*} u &= x^{2} \\ v &= (1 + x)^{\frac{1}{2}} \end{align*} \] ### Step 2: Compute \( \frac{du}{dx} \) and \( \frac{dv}{dx} \) Differentiate \( u \) and \( v \) with respect to \( x \): \[ \begin{align*} \frac{du}{dx} &= \frac{d}{dx} \left( x^{2} \right) = 2x \\ \frac{dv}{dx} &= \frac{d}{dx} \left( (1 + x)^{\frac{1}{2}} \right) = \frac{1}{2}(1 + x)^{-\frac{1}{2}} \cdot 1 = \frac{1}{2}(1 + x)^{-\frac{1}{2}} \end{align*} \] ### Step 3: Apply the Product Rule Substitute \( u \), \( \frac{du}{dx} \), \( v \), and \( \frac{dv}{dx} \) into the Product Rule formula: \[ \frac{dy}{dx} = u \cdot \frac{dv}{dx} + v \cdot \frac{du}{dx} \] Plugging in the values: \[ \frac{dy}{dx} = x^{2} \cdot \left( \frac{1}{2}(1 + x)^{-\frac{1}{2}} \right) + (1 + x)^{\frac{1}{2}} \cdot 2x \] ### Step 4: Simplify the Expression Let's simplify each term: 1. **First Term:** \[ x^{2} \cdot \frac{1}{2}(1 + x)^{-\frac{1}{2}} = \frac{x^{2}}{2 \sqrt{1 + x}} \] 2. **Second Term:** \[ (1 + x)^{\frac{1}{2}} \cdot 2x = 2x \sqrt{1 + x} \] Putting it all together: \[ \frac{dy}{dx} = \frac{x^{2}}{2 \sqrt{1 + x}} + 2x \sqrt{1 + x} \] ### Step 5: Combine the Terms (Optional) If you want to combine the two terms into a single fraction: \[ \frac{dy}{dx} = \frac{x^{2}}{2 \sqrt{1 + x}} + \frac{4x(1 + x)}{2 \sqrt{1 + x}} = \frac{x^{2} + 4x(1 + x)}{2 \sqrt{1 + x}} \] Simplify the numerator: \[ x^{2} + 4x(1 + x) = x^{2} + 4x + 4x^{2} = 5x^{2} + 4x \] So, the derivative becomes: \[ \frac{dy}{dx} = \frac{5x^{2} + 4x}{2 \sqrt{1 + x}} \] ### Final Answer \[ \boxed{ \frac{dy}{dx} = \frac{x^2}{2\,\sqrt{1 + x}} + 2x\,\sqrt{1 + x} } \]