Differentiate each of the following with respect to x in following question: `sin^(3)x cos^(3)x`

Answer

To differentiate the function \( y = \sin^3 x \cos^3 x \) with respect to \( x \), we will use the product rule of differentiation. The product rule states that if you have two functions \( u \) and \( v \), then the derivative of their product is given by:

\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]

### Step-by-Step Solution:

1. **Identify the functions**:
Let \( u = \sin^3 x \) and \( v = \cos^3 x \).

2. **Differentiate \( u \) and \( v \)**:
- To differentiate \( u = \sin^3 x \), we apply the chain rule:
\[ \frac{du}{dx} = 3 \sin^2 x \cdot \frac{d}{dx}(\sin x) = 3 \sin^2 x \cdot \cos x \]
- To differentiate \( v = \cos^3 x \), we again apply the chain rule:
\[ \frac{dv}{dx} = 3 \cos^2 x \cdot \frac{d}{dx}(\cos x) = 3 \cos^2 x \cdot (-\sin x) = -3 \cos^2 x \sin x \]

3. **Apply the product rule**:
Now, we can substitute \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \) into the product rule formula:
\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]
Substituting the values we have:
\[ \frac{dy}{dx} = \sin^3 x \cdot (-3 \cos^2 x \sin x) + \cos^3 x \cdot (3 \sin^2 x \cos x) \]

4. **Simplify the expression**:
- The first term becomes:
\[ -3 \sin^4 x \cos^2 x \]
- The second term becomes:
\[ 3 \cos^4 x \sin^2 x \]
- Thus, we can combine these results:
\[ \frac{dy}{dx} = -3 \sin^4 x \cos^2 x + 3 \cos^4 x \sin^2 x \]

5. **Factor out common terms**:
We can factor out \( 3 \sin^2 x \cos^2 x \):
\[ \frac{dy}{dx} = 3 \sin^2 x \cos^2 x (\cos^2 x - \sin^2 x) \]

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