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We are given the function \( f(x) = (\cos x)^{\cos x} \), and we need to find \( f'(0) \).

### Step-by-Step Solution:

1. **Rewrite the function in a more manageable form**:
   We can express the function as:
   \[
   f(x) = (\cos x)^{\cos x} = e^{\cos x \cdot \ln(\cos x)}
   \]
   This uses the identity \( a^b = e^{b \ln(a)} \).

2. **Differentiate using the chain rule**:
   To find \( f'(x) \), differentiate \( e^{\cos x \cdot \ln(\cos x)} \):
   \[
   f'(x) = e^{\cos x \cdot \ln(\cos x)} \cdot \frac{d}{dx} \left( \cos x \cdot \ln(\cos x) \right)
   \]

3. **Differentiate \( \cos x \cdot \ln(\cos x) \)**:
   Use the product rule. Let:
   \[
   u = \cos x, \quad v = \ln(\cos x)
   \]
   Then:
   \[
   \frac{d}{dx}(uv) = u'v + uv'
   \]
   - \( u' = -\sin x \)
   - \( v = \ln(\cos x) \)
   - \( v' = \frac{-\sin x}{\cos x} = -\tan x \)

   So:
   \[
   \frac{d}{dx} \left( \cos x \cdot \ln(\cos x) \right) = (-\sin x) \ln(\cos x) + \cos x \cdot (-\tan x)
   \]
   Simplify the second term:
   \[
   \cos x \cdot (-\tan x) = -\sin x
   \]
   Thus:
   \[
   \frac{d}{dx} \left( \cos x \cdot \ln(\cos x) \right) = -\sin x \ln(\cos x) - \sin x = -\sin x (\ln(\cos x) + 1)
   \]

4. **Evaluate at \( x = 0 \)**:
   At \( x = 0 \), we have:
   - \( \cos(0) = 1 \)
   - \( \sin(0) = 0 \)
   - \( \ln(\cos(0)) = \ln(1) = 0 \)

   So, the derivative becomes:
   \[
   f'(0) = e^{0} \cdot (-0(0 + 1)) = 1 \cdot 0 = 0
   \]

Thus, \( f'(0) = 0 \).

The correct answer is **(d) zero**.

---

Do you want more details or have any questions?

Here are 5 related questions:
1. How do you differentiate logarithmic functions involving trigonometric expressions?
2. What is the chain rule for derivatives, and how does it apply in this case?
3. Why does the product rule apply when differentiating \( \cos x \cdot \ln(\cos x) \)?
4. How can we evaluate limits or derivatives when logarithmic functions are involved at special points like \( x = 0 \)?
5. Can we generalize the method used here to other functions of the form \( (\text{trig}(x))^{\text{trig}(x)} \)?

**Tip**: Always check if logarithmic simplifications can help you differentiate complex expressions more easily.

If f(x) = (cos(x))^cos(x), then f'(zero) = ?

Find the derivative of f(x) = (cos(x))^cos(x) at x = 0 using the chain rule, product rule, and logarithmic differentiation. Detailed step-by-step solution for students in Grades 11-12 studying calculus.

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