Differentiate the function with respect to x : `2^(cos^(2)x) `

Explanation

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

### Step 1: Rewrite the function
Let \( y = 2^{\cos^2 x} \).

### Step 2: Take the logarithm of both sides
Taking the natural logarithm on both sides gives us:
\[ \log y = \log(2^{\cos^2 x}) \]
Using the property of logarithms, we can simplify the right side:
\[ \log y = \cos^2 x \cdot \log 2 \]

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

### Step 4: Differentiate \( \cos^2 x \)
Using the chain rule again on \( \cos^2 x \):
\[ \frac{d}{dx}(\cos^2 x) = 2 \cos x \cdot \frac{d}{dx}(\cos x) = 2 \cos x \cdot (-\sin x) = -2 \sin x \cos x \]
So we have:
\[ \frac{1}{y} \frac{dy}{dx} = \log 2 \cdot (-2 \sin x \cos x) \]

### Step 5: Solve for \( \frac{dy}{dx} \)
Now we can solve for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = -2 \sin x \cos x \cdot y \cdot \log 2 \]

### Step 6: Substitute back for \( y \)
Recall that \( y = 2^{\cos^2 x} \), so substituting back gives us:
\[ \frac{dy}{dx} = -2 \sin x \cos x \cdot 2^{\cos^2 x} \cdot \log 2 \]

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

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