Find f'(x) when
`f(x)=2^(cos^(2)x)`

Answer

To find the derivative \( f'(x) \) of the function \( f(x) = 2^{\cos^2 x} \), we can follow these steps:

### Step 1: Take the natural logarithm of both sides
We start by taking the logarithm of \( f(x) \):
\[ \log f(x) = \log(2^{\cos^2 x}) \]

### Step 2: Use the logarithmic property
Using the property of logarithms, we can simplify the right side:
\[ \log f(x) = \cos^2 x \cdot \log 2 \]

### Step 3: Differentiate both sides
Now we differentiate both sides with respect to \( x \). Using the chain rule on the left side and the product rule on the right side:
\[ \frac{d}{dx}(\log f(x)) = \frac{1}{f(x)} \cdot f'(x) \]
\[ \frac{d}{dx}(\cos^2 x \cdot \log 2) = \log 2 \cdot \frac{d}{dx}(\cos^2 x) \]

### Step 4: Differentiate \( \cos^2 x \)
Using the chain rule, we find the derivative of \( \cos^2 x \):
\[ \frac{d}{dx}(\cos^2 x) = 2 \cos x \cdot (-\sin x) = -2 \cos x \sin x \]

### Step 5: Substitute back into the derivative
Now substituting back, we have:
\[ \frac{1}{f(x)} \cdot f'(x) = \log 2 \cdot (-2 \cos x \sin x) \]

### Step 6: Solve for \( f'(x) \)
Multiplying both sides by \( f(x) \):
\[ f'(x) = f(x) \cdot \log 2 \cdot (-2 \cos x \sin x) \]

### Step 7: Substitute \( f(x) \) back into the equation
Recall that \( f(x) = 2^{\cos^2 x} \):
\[ f'(x) = 2^{\cos^2 x} \cdot \log 2 \cdot (-2 \cos x \sin x) \]

### Step 8: Simplify the expression
We can express \( -2 \cos x \sin x \) as \( -\sin(2x) \):
\[ f'(x) = -2^{\cos^2 x} \cdot \log 2 \cdot \sin(2x) \]

### Final Answer
Thus, the derivative \( f'(x) \) is:
\[ f'(x) = -\log 2 \cdot 2^{\cos^2 x} \cdot \sin(2x) \]

---