Differentiate the following w.r.t. x :
`x^(2)e^(x)sinx`

Answer

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

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

In our case, we have three functions multiplied together: \( u = x^2 \), \( v = e^x \), and \( w = \sin x \). We will apply the product rule step by step.

### Step 1: Identify the functions
Let:
- \( u = x^2 \)
- \( v = e^x \)
- \( w = \sin x \)

### Step 2: Differentiate each function
Now, we find the derivatives of each function:
- \( u' = \frac{d}{dx}(x^2) = 2x \)
- \( v' = \frac{d}{dx}(e^x) = e^x \)
- \( w' = \frac{d}{dx}(\sin x) = \cos x \)

### Step 3: Apply the product rule
Using the product rule for three functions, we have:

\[ \frac{dy}{dx} = u'vw + uv'w + uvw' \]

Substituting the values we have:

\[ \frac{dy}{dx} = (2x)(e^x)(\sin x) + (x^2)(e^x)(\sin x) + (x^2)(e^x)(\cos x) \]

### Step 4: Simplify the expression
Now we can factor out the common terms:

\[ \frac{dy}{dx} = e^x \left( 2x \sin x + x^2 \sin x + x^2 \cos x \right) \]

### Final Result
Thus, the derivative of \( y = x^2 e^x \sin x \) with respect to \( x \) is:

\[ \frac{dy}{dx} = e^x \left( 2x \sin x + x^2 \sin x + x^2 \cos x \right) \]

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