Find `dy/dx" if "(sinx)^(cosy)=(cosy)^(sinx)`.

Answer

To find \(\frac{dy}{dx}\) for the equation \((\sin x)^{\cos y} = (\cos y)^{\sin x}\), we will use implicit differentiation. Here’s the step-by-step solution:

### Step 1: Take the natural logarithm of both sides
We start by taking the natural logarithm of both sides of the equation to simplify the powers:

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

### Step 2: Apply the logarithmic identity
Using the property of logarithms, \(\ln(a^b) = b \ln a\), we can rewrite the equation:

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

### Step 3: Differentiate both sides with respect to \(x\)
Now we differentiate both sides with respect to \(x\). We will use the product rule and chain rule where necessary.

**Left Side:**
Using the product rule on the left side:

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

The derivative of \(\ln(\sin x)\) is \(\frac{\cos x}{\sin x} = \cot x\). Thus, we have:

\[ \frac{d}{dx}(\cos y) \cdot \ln(\sin x) + \cos y \cdot \cot x \]

Using the chain rule, \(\frac{d(\cos y)}{dx} = -\sin y \cdot \frac{dy}{dx}\):

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

**Right Side:**
Now for the right side:

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

The derivative of \(\ln(\cos y)\) is \(-\tan y \cdot \frac{dy}{dx}\):

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

### Step 4: Set the derivatives equal
Now we set the derivatives from both sides equal to each other:

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

### Step 5: Collect all \(\frac{dy}{dx}\) terms
Rearranging the equation to isolate \(\frac{dy}{dx}\):

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

Factoring out \(\frac{dy}{dx}\):

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

### Step 6: Solve for \(\frac{dy}{dx}\)
Finally, we solve for \(\frac{dy}{dx}\):

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

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

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