Differentiate `2^(cos^(2)x) `

Explanation

To differentiate the function \( y = 2^{\cos^2 x} \), we can follow these steps:

### Step 1: Take the natural logarithm of both sides
We start by taking the logarithm of both sides to simplify the differentiation process:
\[ \ln y = \ln(2^{\cos^2 x}) \]

### Step 2: Use the property of logarithms
Using the property of logarithms that states \( \ln(a^b) = b \ln a \), we can rewrite the equation:
\[ \ln y = \cos^2 x \cdot \ln 2 \]

### Step 3: Differentiate both sides with respect to \( x \)
Now we differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(\ln y) = \frac{d}{dx}(\cos^2 x \cdot \ln 2) \]

### Step 4: Apply the chain rule on the left side
Using the chain rule on the left side, we have:
\[ \frac{1}{y} \frac{dy}{dx} = \ln 2 \cdot \frac{d}{dx}(\cos^2 x) \]

### Step 5: Differentiate \( \cos^2 x \)
Now, we differentiate \( \cos^2 x \) using the chain rule:
\[ \frac{d}{dx}(\cos^2 x) = 2 \cos x \cdot (-\sin x) = -2 \cos x \sin x \]

### Step 6: Substitute back into the equation
Substituting this back into our equation gives:
\[ \frac{1}{y} \frac{dy}{dx} = \ln 2 \cdot (-2 \cos x \sin x) \]

### Step 7: Solve for \( \frac{dy}{dx} \)
Now we multiply both sides by \( y \) to solve for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = y \cdot \ln 2 \cdot (-2 \cos x \sin x) \]

### Step 8: Substitute \( y \) back in
Recall that \( y = 2^{\cos^2 x} \):
\[ \frac{dy}{dx} = 2^{\cos^2 x} \cdot \ln 2 \cdot (-2 \cos x \sin x) \]

### Step 9: Simplify the expression
Using the identity \( 2 \cos x \sin x = \sin(2x) \):
\[ \frac{dy}{dx} = -2^{\cos^2 x} \cdot \ln 2 \cdot \sin(2x) \]

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