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

Answer

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

### Step 1: Define the function
Let \( y = \cos(x^x) \).

### Step 2: Introduce a substitution
Let \( \theta = x^x \). Therefore, we can rewrite the function as \( y = \cos(\theta) \).

### Step 3: Differentiate \( \theta \) with respect to \( x \)
To differentiate \( \theta = x^x \), we first take the natural logarithm of both sides:
\[ \ln(\theta) = \ln(x^x) \]
Using the property of logarithms, we can simplify this:
\[ \ln(\theta) = x \ln(x) \]

### Step 4: Differentiate both sides with respect to \( x \)
Now we differentiate both sides:
\[ \frac{d}{dx}(\ln(\theta)) = \frac{d}{dx}(x \ln(x)) \]
Using the chain rule on the left side and the product rule on the right side:
\[ \frac{1}{\theta} \frac{d\theta}{dx} = \ln(x) + 1 \]
Thus, we can express \( \frac{d\theta}{dx} \):
\[ \frac{d\theta}{dx} = \theta (\ln(x) + 1) \]
Substituting back \( \theta = x^x \):
\[ \frac{d\theta}{dx} = x^x (\ln(x) + 1) \]

### Step 5: Differentiate \( y \) with respect to \( x \)
Now we differentiate \( y = \cos(\theta) \):
\[ \frac{dy}{dx} = -\sin(\theta) \frac{d\theta}{dx} \]
Substituting \( \frac{d\theta}{dx} \):
\[ \frac{dy}{dx} = -\sin(\theta) \cdot x^x (\ln(x) + 1) \]

### Step 6: Substitute back \( \theta \)
Finally, we substitute \( \theta = x^x \) back into the equation:
\[ \frac{dy}{dx} = -\sin(x^x) \cdot x^x (\ln(x) + 1) \]

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

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