Find dy/dx, if `y=sin^3 x cosx`

Options

A. `sin^2 x(3 cos^2 x-sin^2 x)`
B. `3cos^3 (1-sin^2 x)`
C. `3 cos^4 x`
D. `3sin^4 x`

Answer

To find \( \frac{dy}{dx} \) for the function \( y = \sin^3 x \cos x \), we will use the product rule and the chain rule of differentiation.

### Step-by-Step Solution:

1. **Identify the function**:
We have \( y = \sin^3 x \cos x \). This is a product of two functions: \( u = \sin^3 x \) and \( v = \cos x \).

2. **Apply the Product Rule**:
The product rule states that if \( y = uv \), then \( \frac{dy}{dx} = u'v + uv' \).
Here, \( u = \sin^3 x \) and \( v = \cos x \).

3. **Differentiate \( u \)**:
To differentiate \( u = \sin^3 x \), we will use the chain rule:
\[ u' = 3 \sin^2 x \cdot \frac{d}{dx}(\sin x) = 3 \sin^2 x \cos x \]

4. **Differentiate \( v \)**:
The derivative of \( v = \cos x \) is:
\[ v' = -\sin x \]

5. **Substitute into the Product Rule**:
Now we substitute \( u' \), \( v \), \( u \), and \( v' \) into the product rule:
\[ \frac{dy}{dx} = u'v + uv' = (3 \sin^2 x \cos x)(\cos x) + (\sin^3 x)(-\sin x) \]

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

7. **Final result**:
Therefore, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = 3 \sin^2 x \cos^2 x - \sin^4 x \]