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 three functions \( u, v, w \), then the derivative is given by:

\[ \frac{d}{dx}(uvw) = u'vw + uv'w + uvw' \]

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

Now, we will differentiate each of these functions:

1. **Differentiate \( u = x^2 \)**:
\[ u' = \frac{d}{dx}(x^2) = 2x \]

2. **Differentiate \( v = e^x \)**:
\[ v' = \frac{d}{dx}(e^x) = e^x \]

3. **Differentiate \( w = \sin x \)**:
\[ w' = \frac{d}{dx}(\sin x) = \cos x \]

Now, applying the product rule:

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

Substituting the derivatives and the original functions:

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

This simplifies to:

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

Now, we can factor out the common terms \( e^x \) and \( \sin x \):

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

Combining the terms inside the brackets:

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

Thus, the final result is:

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