`(d)/(dx)(cosx-(1)/(3)cos^(3)x)=`

Options

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

Answer

To solve the problem \(\frac{d}{dx}\left(\cos x - \frac{1}{3}\cos^3 x\right)\), we will differentiate each term step by step.

### Step-by-Step Solution:

1. **Identify the function to differentiate:**
\[ y = \cos x - \frac{1}{3}\cos^3 x \]

2. **Differentiate the first term:**
The derivative of \(\cos x\) is:
\[ \frac{d}{dx}(\cos x) = -\sin x \]

3. **Differentiate the second term using the chain rule:**
For the term \(-\frac{1}{3}\cos^3 x\), we apply the chain rule. Let \(u = \cos x\), then:
\[ \frac{d}{dx}\left(-\frac{1}{3}u^3\right) = -\frac{1}{3} \cdot 3u^2 \cdot \frac{du}{dx} \]
Here, \(\frac{du}{dx} = -\sin x\), so:
\[ \frac{d}{dx}\left(-\frac{1}{3}\cos^3 x\right) = -\frac{1}{3} \cdot 3\cos^2 x \cdot (-\sin x) = \cos^2 x \sin x \]

4. **Combine the derivatives:**
Now, we combine the derivatives of both terms:
\[ \frac{dy}{dx} = -\sin x + \cos^2 x \sin x \]

5. **Factor out \(-\sin x\):**
We can factor out \(-\sin x\) from the expression:
\[ \frac{dy}{dx} = -\sin x (1 - \cos^2 x) \]

6. **Use the Pythagorean identity:**
Recall that \(1 - \cos^2 x = \sin^2 x\):
\[ \frac{dy}{dx} = -\sin x \cdot \sin^2 x \]

7. **Final result:**
Thus, we have:
\[ \frac{dy}{dx} = -\sin^3 x \]

### Final Answer:
\[ \frac{d}{dx}\left(\cos x - \frac{1}{3}\cos^3 x\right) = -\sin^3 x \]