Find `(dy)/(dx)` for the functions: `y=x^3e^xsinx`

Explanation

To find \(\frac{dy}{dx}\) for the function \(y = x^3 e^x \sin x\), we will use the product rule of differentiation. The product rule states that if you have a product of functions \(u\), \(v\), and \(w\), then:

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

In our case, we can let:
- \(u = x^3\)
- \(v = e^x\)
- \(w = \sin x\)

Now, we will differentiate each of these functions:

1. **Differentiate \(u = x^3\)**:
\[ u' = \frac{d}{dx}(x^3) = 3x^2 \]

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, we can apply the product rule:

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

Substituting the derivatives and the original functions:

\[ \frac{dy}{dx} = (3x^2)(e^x)(\sin x) + (x^3)(e^x)(\sin x) + (x^3)(e^x)(\cos x) \]

Now we can simplify this expression:

\[ \frac{dy}{dx} = 3x^2 e^x \sin x + x^3 e^x \sin x + x^3 e^x \cos x \]

Next, we can factor out the common terms \(e^x\) and \(x^2\):

\[ \frac{dy}{dx} = e^x x^2 \left( 3 \sin x + x \sin x + x \cos x \right) \]

Thus, the final result is:

\[ \frac{dy}{dx} = e^x x^2 \left( (3 + x) \sin x + x \cos x \right) \]