Differentiate `(xcosx)^x` with respect to `xdot`

Explanation

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

### Step 1: Take the natural logarithm of both sides
Let \( y = (x \cos x)^x \).

Taking the natural logarithm of both sides, we have:
\[ \ln y = \ln((x \cos x)^x) \]

### Step 2: Simplify using logarithmic properties
Using the property of logarithms \( \ln(a^b) = b \ln a \), we can simplify:
\[ \ln y = x \ln(x \cos x) \]

### Step 3: Expand the logarithm
Using the property \( \ln(ab) = \ln a + \ln b \):
\[ \ln y = x (\ln x + \ln(\cos x)) \]
This simplifies to:
\[ \ln y = x \ln x + x \ln(\cos x) \]

### Step 4: Differentiate both sides with respect to \( x \)
Now we differentiate both sides using implicit differentiation:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(x \ln x) + \frac{d}{dx}(x \ln(\cos x)) \]

### Step 5: Apply the product rule
For the first term \( x \ln x \):
\[ \frac{d}{dx}(x \ln x) = \ln x + 1 \]

For the second term \( x \ln(\cos x) \), we apply the product rule:
\[ \frac{d}{dx}(x \ln(\cos x)) = \ln(\cos x) + x \frac{d}{dx}(\ln(\cos x)) \]
Using the chain rule, we find:
\[ \frac{d}{dx}(\ln(\cos x)) = -\tan x \]
Thus,
\[ \frac{d}{dx}(x \ln(\cos x)) = \ln(\cos x) - x \tan x \]

### Step 6: Combine the results
Combining both derivatives, we have:
\[ \frac{1}{y} \frac{dy}{dx} = (\ln x + 1) + (\ln(\cos x) - x \tan x) \]
This simplifies to:
\[ \frac{1}{y} \frac{dy}{dx} = \ln x + 1 + \ln(\cos x) - x \tan x \]

### Step 7: Solve for \( \frac{dy}{dx} \)
Multiplying both sides by \( y \):
\[ \frac{dy}{dx} = y \left( \ln x + 1 + \ln(\cos x) - x \tan x \right) \]

### Step 8: Substitute back for \( y \)
Substituting back \( y = (x \cos x)^x \):
\[ \frac{dy}{dx} = (x \cos x)^x \left( \ln x + 1 + \ln(\cos x) - x \tan x \right) \]

### Final Answer
Thus, the derivative of \( (x \cos x)^x \) with respect to \( x \) is:
\[ \frac{dy}{dx} = (x \cos x)^x \left( \ln x + 1 + \ln(\cos x) - x \tan x \right) \]