Find `(dy)/(dx)`, when:
`y=(cosx)^(cosx)`

Answer

To find \(\frac{dy}{dx}\) when \(y = (\cos x)^{\cos x}\), we can follow these steps:

### Step 1: Take the natural logarithm of both sides
We start by taking the logarithm of both sides to simplify the differentiation process.

\[ \ln y = \ln((\cos x)^{\cos x}) \]

### Step 2: Use the property of logarithms
Using the property of logarithms that states \(\ln(a^b) = b \cdot \ln a\), we can rewrite the equation:

\[ \ln y = \cos x \cdot \ln(\cos x) \]

### Step 3: Differentiate both sides
Now we differentiate both sides with respect to \(x\). We will use implicit differentiation on the left side and the product rule on the right side.

\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\cos x \cdot \ln(\cos x)) \]

Using the product rule, we have:

\[ \frac{d}{dx}(\cos x \cdot \ln(\cos x)) = \frac{d}{dx}(\cos x) \cdot \ln(\cos x) + \cos x \cdot \frac{d}{dx}(\ln(\cos x)) \]

### Step 4: Differentiate \(\cos x\) and \(\ln(\cos x)\)
Now we differentiate each part:

1. The derivative of \(\cos x\) is \(-\sin x\).
2. The derivative of \(\ln(\cos x)\) using the chain rule is \(\frac{-\sin x}{\cos x} = -\tan x\).

Putting these together, we have:

\[ \frac{1}{y} \frac{dy}{dx} = (-\sin x) \cdot \ln(\cos x) + \cos x \cdot (-\tan x) \]

### Step 5: Simplify the right-hand side
The term \(\cos x \cdot (-\tan x)\) can be rewritten as \(-\sin x\):

\[ \frac{1}{y} \frac{dy}{dx} = -\sin x \ln(\cos x) - \sin x \]

### Step 6: Multiply both sides by \(y\)
Now, we multiply both sides by \(y\) to isolate \(\frac{dy}{dx}\):

\[ \frac{dy}{dx} = y \left(-\sin x \ln(\cos x) - \sin x\right) \]

### Step 7: Substitute back for \(y\)
Recall that \(y = (\cos x)^{\cos x}\):

\[ \frac{dy}{dx} = (\cos x)^{\cos x} \left(-\sin x \ln(\cos x) - \sin x\right) \]

### Final Result
Thus, the derivative \(\frac{dy}{dx}\) is:

\[ \frac{dy}{dx} = -\sin x (\cos x)^{\cos x} \left(\ln(\cos x) + 1\right) \]