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

Answer

To differentiate the function \( y = (\log x)^{\cos x} \) with respect to \( x \), we will use logarithmic differentiation. Here are the steps:

### Step 1: Take the natural logarithm of both sides
We start by taking the natural logarithm of both sides:
\[ \log y = \log((\log x)^{\cos x}) \]

### Step 2: Simplify using logarithmic properties
Using the property of logarithms that states \( \log(a^b) = b \log a \), we can simplify the right-hand side:
\[ \log y = \cos x \cdot \log(\log x) \]

### Step 3: Differentiate both sides
Now, we differentiate both sides with respect to \( x \). Remember to use the chain rule on the left side and the product rule on the right side:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\cos x \cdot \log(\log x)) \]

Using the product rule on the right side:
\[ \frac{d}{dx}(\cos x \cdot \log(\log x)) = \frac{d}{dx}(\cos x) \cdot \log(\log x) + \cos x \cdot \frac{d}{dx}(\log(\log x)) \]
\[ = -\sin x \cdot \log(\log x) + \cos x \cdot \frac{1}{\log x} \cdot \frac{1}{x} \]

### Step 4: Substitute back to find \( \frac{dy}{dx} \)
Now we can substitute back to find \( \frac{dy}{dx} \):
\[ \frac{1}{y} \frac{dy}{dx} = -\sin x \cdot \log(\log x) + \frac{\cos x}{x \log x} \]
Multiplying both sides by \( y \):
\[ \frac{dy}{dx} = y \left(-\sin x \cdot \log(\log x) + \frac{\cos x}{x \log x}\right) \]

### Step 5: Substitute \( y \) back in
Since \( y = (\log x)^{\cos x} \), we substitute back:
\[ \frac{dy}{dx} = (\log x)^{\cos x} \left(-\sin x \cdot \log(\log x) + \frac{\cos x}{x \log x}\right) \]

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

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