If `(cosx)^(y)=(siny)^(x),` then find `(dy)/(dx)`.

Answer

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

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

### Step 2: Apply the logarithmic power rule
Using the property of logarithms that allows us to bring the exponent down, we rewrite the equation as:
\[ y \log(\cos x) = x \log(\sin y) \]

### 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 for differentiation:
\[ \frac{d}{dx}(y \log(\cos x)) = \frac{d}{dx}(x \log(\sin y)) \]

Using the product rule on the left side:
\[ \frac{dy}{dx} \log(\cos x) + y \frac{d}{dx}(\log(\cos x)) = \frac{d}{dx}(x \log(\sin y)) \]

For the left side, we differentiate \(\log(\cos x)\):
\[ \frac{d}{dx}(\log(\cos x)) = \frac{-\sin x}{\cos x} = -\tan x \]
Thus, the left side becomes:
\[ \frac{dy}{dx} \log(\cos x) - y \tan x \]

For the right side, we apply the product rule:
\[ \frac{d}{dx}(x \log(\sin y)) = \log(\sin y) + x \frac{d}{dx}(\log(\sin y)) \]
Using the chain rule on \(\log(\sin y)\):
\[ \frac{d}{dx}(\log(\sin y)) = \frac{\cos y}{\sin y} \frac{dy}{dx} = \cot y \frac{dy}{dx} \]
Thus, the right side becomes:
\[ \log(\sin y) + x \cot y \frac{dy}{dx} \]

### Step 4: Set the derivatives equal
Now we have:
\[ \frac{dy}{dx} \log(\cos x) - y \tan x = \log(\sin y) + x \cot y \frac{dy}{dx} \]

### Step 5: Collect all \(\frac{dy}{dx}\) terms on one side
Rearranging gives:
\[ \frac{dy}{dx} \log(\cos x) - x \cot y \frac{dy}{dx} = \log(\sin y) + y \tan x \]
Factoring out \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} \left(\log(\cos x) - x \cot y\right) = \log(\sin y) + y \tan x \]

### Step 6: Solve for \(\frac{dy}{dx}\)
Finally, we isolate \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{\log(\sin y) + y \tan x}{\log(\cos x) - x \cot y} \]

### Final Answer:
\[ \frac{dy}{dx} = \frac{\log(\sin y) + y \tan x}{\log(\cos x) - x \cot y} \]