`(sinx)^(cosx)`

Explanation

To find the differentiation of the function \( y = (\sin x)^{\cos x} \), we will follow these steps:

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

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

### Step 3: Differentiate both sides
Now we differentiate both sides with respect to \( x \). Using implicit differentiation on the left side and the product rule on the right side:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\cos x \cdot \log(\sin x)) \]

### Step 4: Apply the product rule
Let \( u = \cos x \) and \( v = \log(\sin x) \). By the product rule:
\[ \frac{d}{dx}(uv) = u'v + uv' \]
where \( u' = -\sin x \) and \( v' = \frac{1}{\sin x} \cdot \cos x = \cot x \).

Thus, we have:
\[ \frac{1}{y} \frac{dy}{dx} = (-\sin x) \log(\sin x) + \cos x \cdot \cot x \]

### Step 5: Solve for \( \frac{dy}{dx} \)
Now, we can multiply both sides by \( y \) to isolate \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = y \left( -\sin x \log(\sin x) + \cos x \cot x \right) \]

### Step 6: Substitute back for \( y \)
Since we know \( y = (\sin x)^{\cos x} \), we substitute back:
\[ \frac{dy}{dx} = (\sin x)^{\cos x} \left( -\sin x \log(\sin x) + \cos x \cot x \right) \]

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

---