Differentiate `(sinx)^(cosx)`

Explanation

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

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

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

### Step 3: Differentiate both sides with respect to \( x \)
Now we differentiate both sides with respect to \( x \). For the left side, we use the chain rule:
\[ \frac{d}{dx}(\ln y) = \frac{1}{y} \frac{dy}{dx} \]
For the right side, we apply the product rule:
\[ \frac{d}{dx}(\cos x \cdot \ln(\sin x)) = \frac{d}{dx}(\cos x) \cdot \ln(\sin x) + \cos x \cdot \frac{d}{dx}(\ln(\sin x)) \]

### Step 4: Differentiate the components
Now we differentiate each component:
- The derivative of \( \cos x \) is \( -\sin x \).
- The derivative of \( \ln(\sin x) \) is \( \frac{1}{\sin x} \cdot \cos x \) (using the chain rule).

Putting these together, we have:
\[ \frac{dy}{dx} \cdot \frac{1}{y} = -\sin x \cdot \ln(\sin x) + \cos x \cdot \frac{\cos x}{\sin x} \]

### Step 5: Simplify the right side
This simplifies to:
\[ \frac{dy}{dx} \cdot \frac{1}{y} = -\sin x \cdot \ln(\sin x) + \frac{\cos^2 x}{\sin x} \]

### Step 6: Multiply both sides by \( y \)
Now, we multiply both sides by \( y \) to isolate \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = y \left( -\sin x \cdot \ln(\sin x) + \frac{\cos^2 x}{\sin x} \right) \]

### Step 7: Substitute back for \( y \)
Since \( y = (\sin x)^{\cos x} \), we substitute back:
\[ \frac{dy}{dx} = (\sin x)^{\cos x} \left( -\sin x \cdot \ln(\sin x) + \frac{\cos^2 x}{\sin x} \right) \]

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