If `(sinx)^(y)=(siny)^(x),"find "dy/dx`.

Answer

To solve the equation \((\sin x)^y = (\sin y)^x\) and find \(\frac{dy}{dx}\), we can follow these steps:

### Step 1: Take the logarithm of both sides
We start with the equation:
\[ (\sin x)^y = (\sin y)^x \]
Taking the natural logarithm on both sides gives:
\[ \log((\sin x)^y) = \log((\sin y)^x) \]

### Step 2: Apply the logarithmic power rule
Using the property of logarithms that states \(\log(a^b) = b \log(a)\), we can rewrite the equation as:
\[ y \log(\sin x) = x \log(\sin y) \]

### Step 3: Differentiate both sides with respect to \(x\)
Now we differentiate both sides with respect to \(x\):
\[ \frac{d}{dx}(y \log(\sin x)) = \frac{d}{dx}(x \log(\sin y)) \]

### Step 4: Apply the product rule
Using the product rule \(\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}\), we differentiate the left side:
\[ \frac{dy}{dx} \log(\sin x) + y \frac{d}{dx}(\log(\sin x)) \]
The derivative of \(\log(\sin x)\) is:
\[ \frac{1}{\sin x} \cos x = \cot x \]
So the left side becomes:
\[ \frac{dy}{dx} \log(\sin x) + y \cot x \]

For the right side:
\[ \frac{d}{dx}(x \log(\sin y)) = \log(\sin y) + x \frac{d}{dx}(\log(\sin y)) \]
The derivative of \(\log(\sin y)\) is:
\[ \frac{1}{\sin y} \cos y \frac{dy}{dx} = \cot y \frac{dy}{dx} \]
So the right side becomes:
\[ \log(\sin y) + x \cot y \frac{dy}{dx} \]

### Step 5: Set the derivatives equal
Now we set both sides equal:
\[ \frac{dy}{dx} \log(\sin x) + y \cot x = \log(\sin y) + x \cot y \frac{dy}{dx} \]

### Step 6: Collect all \(\frac{dy}{dx}\) terms
Rearranging gives:
\[ \frac{dy}{dx} \log(\sin x) - x \cot y \frac{dy}{dx} = \log(\sin y) - y \cot x \]
Factoring out \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} (\log(\sin x) - x \cot y) = \log(\sin y) - y \cot x \]

### Step 7: Solve for \(\frac{dy}{dx}\)
Finally, we can solve for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{\log(\sin y) - y \cot x}{\log(\sin x) - x \cot y} \]

### Final Answer
Thus, the derivative \(\frac{dy}{dx}\) is given by:
\[ \frac{dy}{dx} = \frac{\log(\sin y) - y \cot x}{\log(\sin x) - x \cot y} \]