If `y=sin^(2)x.cos^(3)x,` then `(dy)/(dx)`

Answer

To find the derivative of the function \( y = \sin^2 x \cdot \cos^3 x \), we will use the product rule of differentiation. The product rule states that if you have two functions \( u(x) \) and \( v(x) \), then the derivative of their product is given by:

\[ \frac{d(uv)}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]

### Step-by-Step Solution

1. **Identify the Functions**:
Let \( u = \sin^2 x \) and \( v = \cos^3 x \).

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

3. **Apply the Product Rule**:
Now, using the product rule:
\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]
Substituting \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \):
\[ \frac{dy}{dx} = \sin^2 x \cdot (-3 \cos^2 x \sin x) + \cos^3 x \cdot (2 \sin x \cos x) \]

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

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

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