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 properties 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 \( x \). Remember to use the chain rule on the left side:
\[ \frac{1}{y} \frac{dy}{dx} = \ln 2 \cdot \frac{d}{dx}(\cos^2 x) \]

### Step 4: Differentiate \( \cos^2 x \)
To differentiate \( \cos^2 x \), we use the chain rule:
\[ \frac{d}{dx}(\cos^2 x) = 2 \cos x \cdot (-\sin x) = -2 \cos x \sin x = -\sin(2x) \]
(using the double angle identity for sine).

### Step 5: Substitute back into the equation
Substituting this back into our differentiation equation gives:
\[ \frac{1}{y} \frac{dy}{dx} = \ln 2 \cdot (-\sin(2x)) \]

### Step 6: Solve for \( \frac{dy}{dx} \)
Now, we multiply both sides by \( y \):
\[ \frac{dy}{dx} = y \cdot \ln 2 \cdot (-\sin(2x)) \]

### Step 7: Substitute back the value of \( y \)
Recall that \( y = 2^{\cos^2 x} \). Substituting this back in gives:
\[ \frac{dy}{dx} = 2^{\cos^2 x} \cdot \ln 2 \cdot (-\sin(2x)) \]

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

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