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

Answer

To find \(\frac{dy}{dx}\) for the function \(y = (\log x)^{\sin x}\), we can follow these steps:

### Step 1: Take the natural logarithm of both sides
We start by taking the natural logarithm of both sides to simplify the differentiation process.
\[ \log y = \sin x \cdot \log(\log x) \]

### Step 2: Differentiate both sides with respect to \(x\)
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.
\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\sin x \cdot \log(\log x)) \]
Using the product rule on the right side:
\[ \frac{d}{dx}(\sin x \cdot \log(\log x)) = \cos x \cdot \log(\log x) + \sin x \cdot \frac{1}{\log x} \cdot \frac{1}{x} \]
This simplifies to:
\[ \frac{d}{dx}(\sin x \cdot \log(\log x)) = \cos x \cdot \log(\log x) + \frac{\sin x}{x \log x} \]

### Step 3: Substitute back into the equation
Now we substitute this back into our differentiated equation:
\[ \frac{1}{y} \frac{dy}{dx} = \cos x \cdot \log(\log x) + \frac{\sin x}{x \log x} \]

### Step 4: Solve for \(\frac{dy}{dx}\)
To isolate \(\frac{dy}{dx}\), we multiply both sides by \(y\):
\[ \frac{dy}{dx} = y \left( \cos x \cdot \log(\log x) + \frac{\sin x}{x \log x} \right) \]

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

### Final Result
Thus, the derivative \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = (\log x)^{\sin x} \left( \cos x \cdot \log(\log x) + \frac{\sin x}{x \log x} \right) \]

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