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

Answer

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

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

\[ \log((\cos x)^y) = \log((\sin y)^x) \]

### Step 2: Apply the power rule of logarithms
Using the property of logarithms that states \( \log(a^b) = b \log(a) \), we can 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 \( x \):

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

### Step 4: Apply the product rule
Using the product rule \( \frac{d}{dx}(u \cdot v) = u \frac{dv}{dx} + v \frac{du}{dx} \), we differentiate the left side:

\[ y \frac{d}{dx}(\log(\cos x)) + \log(\cos x) \frac{dy}{dx} \]

And for the right side:

\[ \log(\sin y) + x \frac{d}{dx}(\log(\sin y)) \]

### Step 5: Differentiate the logarithmic functions
Using the derivatives \( \frac{d}{dx}(\log(u)) = \frac{1}{u} \frac{du}{dx} \):

- For \( \log(\cos x) \):
\[ \frac{d}{dx}(\log(\cos x)) = \frac{-\sin x}{\cos x} = -\tan x \]

- For \( \log(\sin y) \):
\[ \frac{d}{dx}(\log(\sin y)) = \frac{1}{\sin y} \cdot \cos y \cdot \frac{dy}{dx} \]

### Step 6: Substitute the derivatives back into the equation
Substituting these derivatives back into our differentiated equation gives us:

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

### Step 7: Rearrange the equation to isolate \( \frac{dy}{dx} \)
Rearranging the equation to isolate \( \frac{dy}{dx} \):

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

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

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

### Step 8: Solve for \( \frac{dy}{dx} \)
Finally, we solve for \( \frac{dy}{dx} \):

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

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

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