Differentiate the following w.r.t. x :
`e^(x)cos^(3)xsin^(2)x`

Answer

To differentiate the function \( y = e^x \cos^3 x \sin^2 x \) with respect to \( x \), we can use the product rule and chain rule. Here’s a step-by-step solution:

### Step 1: Identify the function
Let:
\[ y = e^x \cos^3 x \sin^2 x \]

### Step 2: Apply the product rule
Since \( y \) is a product of three functions \( u = e^x \), \( v = \cos^3 x \), and \( w = \sin^2 x \), we will use the product rule:
\[ \frac{dy}{dx} = u'vw + uv'w + uvw' \]

### Step 3: Differentiate each part
1. Differentiate \( u = e^x \):
\[ u' = e^x \]

2. Differentiate \( v = \cos^3 x \) using the chain rule:
\[ v' = 3 \cos^2 x (-\sin x) = -3 \cos^2 x \sin x \]

3. Differentiate \( w = \sin^2 x \) using the chain rule:
\[ w' = 2 \sin x \cos x \]

### Step 4: Substitute into the product rule
Now substitute \( u, u', v, v', w, w' \) into the product rule:
\[ \frac{dy}{dx} = e^x \cos^3 x \cdot (2 \sin x \cos x) + e^x (-3 \cos^2 x \sin x) \cdot \sin^2 x + e^x \cos^3 x \cdot (2 \sin x \cos x) \]

### Step 5: Simplify the expression
Combining the terms:
\[ \frac{dy}{dx} = e^x \left( \cos^3 x \cdot (2 \sin x \cos x) - 3 \cos^2 x \sin^3 x + \cos^3 x \cdot (2 \sin x \cos x) \right) \]
\[ = e^x \left( 2 \cos^4 x \sin x - 3 \cos^2 x \sin^3 x \right) \]

### Final Answer
Thus, the derivative of \( y \) with respect to \( x \) is:
\[ \frac{dy}{dx} = e^x \left( 2 \cos^4 x \sin x - 3 \cos^2 x \sin^3 x \right) \]