Differentiate w.r.t x
(i) `y=e^(x)sin^(3)xcos^(4)x` (ii) `y=x*e^(xsinx)`

Answer

To differentiate the given functions with respect to \( x \), we will apply the product rule and chain rule as necessary.

### Part (i): Differentiate \( y = e^{x} \sin^{3} x \cos^{4} x \)

1. **Identify the components**:
We can see that \( y \) is a product of three functions:
- \( f(x) = e^{x} \)
- \( g(x) = \sin^{3} x \)
- \( h(x) = \cos^{4} x \)

2. **Apply the product rule**:
The product rule states that if \( y = u \cdot v \cdot w \), then:
\[ \frac{dy}{dx} = u'vw + uv'w + uvw' \]
Here, \( u = e^{x} \), \( v = \sin^{3} x \), and \( w = \cos^{4} x \).

3. **Differentiate each function**:
- \( u' = \frac{d}{dx}(e^{x}) = e^{x} \)
- For \( v = \sin^{3} x \):
\[ v' = 3 \sin^{2} x \cos x \quad \text{(using chain rule)} \]
- For \( w = \cos^{4} x \):
\[ w' = -4 \cos^{3} x \sin x \quad \text{(using chain rule)} \]

4. **Substitute into the product rule**:
\[ \frac{dy}{dx} = e^{x} \sin^{3} x \cdot \cos^{4} x + e^{x} \cdot 3 \sin^{2} x \cos x \cdot \cos^{4} x + e^{x} \sin^{3} x \cdot (-4 \cos^{3} x \sin x) \]

5. **Simplify the expression**:
\[ \frac{dy}{dx} = e^{x} \sin^{3} x \cos^{4} x + 3 e^{x} \sin^{2} x \cos^{5} x - 4 e^{x} \sin^{4} x \cos^{3} x \]

6. **Factor out common terms**:
\[ \frac{dy}{dx} = e^{x} \sin^{2} x \cos^{3} x \left( \sin x \cos x + 3 \cos^{2} x - 4 \sin^{2} x \right) \]

### Part (ii): Differentiate \( y = x e^{x \sin x} \)

1. **Identify the components**:
Here, we can see \( y \) is a product of:
- \( f(x) = x \)
- \( g(x) = e^{x \sin x} \)

2. **Apply the product rule**:
\[ \frac{dy}{dx} = f' g + f g' \]

3. **Differentiate each function**:
- \( f' = \frac{d}{dx}(x) = 1 \)
- For \( g = e^{x \sin x} \):
- Using the chain rule:
\[ g' = e^{x \sin x} \cdot \frac{d}{dx}(x \sin x) \]
- Now differentiate \( x \sin x \) using the product rule:
\[ \frac{d}{dx}(x \sin x) = \sin x + x \cos x \]
- Thus,
\[ g' = e^{x \sin x} (\sin x + x \cos x) \]

4. **Substitute into the product rule**:
\[ \frac{dy}{dx} = 1 \cdot e^{x \sin x} + x \cdot e^{x \sin x} (\sin x + x \cos x) \]

5. **Factor out common terms**:
\[ \frac{dy}{dx} = e^{x \sin x} \left( 1 + x (\sin x + x \cos x) \right) \]

### Final Answers:
1. \( \frac{dy}{dx} = e^{x} \sin^{2} x \cos^{3} x \left( \sin x \cos x + 3 \cos^{2} x - 4 \sin^{2} x \right) \)
2. \( \frac{dy}{dx} = e^{x \sin x} \left( 1 + x (\sin x + x \cos x) \right) \)