Differentiate the following w.r.t.x.
`(sinx)^(cosx)`

Answer

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

### Step 1: Take the natural logarithm of both sides.
Let:
\[ y = (\sin x)^{\cos x} \]
Taking the natural logarithm on both sides:
\[ \ln y = \ln((\sin x)^{\cos x}) \]

### Step 2: Apply the logarithmic identity.
Using the property of logarithms \( \ln(a^b) = b \ln a \):
\[ \ln y = \cos x \cdot \ln(\sin x) \]

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

### Step 4: Differentiate the right-hand side.
Now we differentiate each term on the right:
- 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).

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

### Step 5: Simplify the equation.
Now we can simplify the right-hand side:
\[ \frac{1}{y} \frac{dy}{dx} = -\sin x \cdot \ln(\sin x) + \frac{\cos^2 x}{\sin x} \]

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

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

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