Find `(dy)/(dx)`, if `y =sin^3x cos x`.

Options

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

Answer

To find \(\frac{dy}{dx}\) for the function \(y = \sin^3 x \cos x\), 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:

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

### Step-by-Step Solution:

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

2. **Differentiate \(u\)**:
We need to find \(\frac{du}{dx}\). Using the chain rule:
\[ u = \sin^3 x \implies \frac{du}{dx} = 3 \sin^2 x \cdot \frac{d}{dx}(\sin x) = 3 \sin^2 x \cos x \]

3. **Differentiate \(v\)**:
Now, differentiate \(v\):
\[ v = \cos x \implies \frac{dv}{dx} = -\sin x \]

4. **Apply the product rule**:
Now, apply the product rule:
\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]
Substituting the values we found:
\[ \frac{dy}{dx} = \sin^3 x (-\sin x) + \cos x (3 \sin^2 x \cos x) \]

5. **Simplify the expression**:
Simplifying the expression:
\[ \frac{dy}{dx} = -\sin^4 x + 3 \sin^2 x \cos^2 x \]
We can factor out \(\sin^2 x\):
\[ \frac{dy}{dx} = \sin^2 x (3 \cos^2 x - \sin^2 x) \]

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