Evaluate differentiation of y with respect to x :
`y=x^(2)sinx`

Options

A. `x^(2)cosx+2sinx`
B. `x^(3)cosx+2xsinx`
C. `x^(2)cosx+2xsinx`
D. `x^(2)cosx-2xsinx`

Explanation

To evaluate the differentiation of \( y \) with respect to \( x \) for the function \( y = x^2 \sin x \), we will use the product rule of differentiation. The product rule states that if you have two functions multiplied together, say \( u \) and \( v \), then the derivative of their product is given by:

\[ \frac{d(uv)}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]

In our case, we can identify:
- \( u = x^2 \)
- \( v = \sin x \)

Now, we will differentiate \( y \) step by step.

### Step 1: Differentiate \( u \) and \( v \)
1. Differentiate \( u = x^2 \):
\[ \frac{du}{dx} = 2x \]

2. Differentiate \( v = \sin x \):
\[ \frac{dv}{dx} = \cos x \]

### Step 2: Apply the Product Rule
Now, we will apply the product rule:
\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]

Substituting the values we found:
\[ \frac{dy}{dx} = x^2 \cdot \cos x + \sin x \cdot 2x \]

### Step 3: Simplify the Expression
Rearranging the terms, we can write:
\[ \frac{dy}{dx} = x^2 \cos x + 2x \sin x \]

### Final Result
Thus, the derivative of \( y \) with respect to \( x \) is:
\[ \frac{dy}{dx} = x^2 \cos x + 2x \sin x \]

---