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

Answer

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

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

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

### Step 2: Use the power rule of logarithms
Using the power rule of logarithms, we can rewrite the right side:

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

### 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.

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

Using the product rule on the right side:

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

### Step 4: Compute the derivatives
Now, we compute the derivatives:

1. \(\frac{d}{dx}(\log x) = \frac{1}{x}\)
2. For \(\frac{d}{dx}(\log(\cos x))\), we use the chain rule:
\[ \frac{d}{dx}(\log(\cos x)) = \frac{-\sin x}{\cos x} = -\tan x \]

Substituting these derivatives back into our equation gives:

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

### Step 5: Multiply both sides by \(y\)
Now, we multiply both sides by \(y\) to isolate \(\frac{dy}{dx}\):

\[ \frac{dy}{dx} = y \left( \frac{\log(\cos x)}{x} - \log x \cdot \tan x \right) \]

### Step 6: Substitute back for \(y\)
Recall that \(y = (\cos x)^{\log x}\), so we substitute back:

\[ \frac{dy}{dx} = (\cos x)^{\log x} \left( \frac{\log(\cos x)}{x} - \log x \cdot \tan x \right) \]

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

\[ \frac{dy}{dx} = (\cos x)^{\log x} \left( \frac{\log(\cos x)}{x} - \log x \cdot \tan x \right) \]