Find `dy/dx` when `(tanx)^(y)=(tany)^(x)`

Answer

To find \( \frac{dy}{dx} \) when \( (\tan x)^y = (\tan y)^x \), we will follow these steps:

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

### Step 2: Use the logarithmic identity
Using the property of logarithms \( \log(a^b) = b \cdot \log(a) \), we can rewrite the equation as:
\[ y \cdot \log(\tan x) = x \cdot \log(\tan y) \]

### Step 3: Differentiate both sides with respect to \( x \)
Now we differentiate both sides with respect to \( x \). We will use the product rule on both sides:
\[ \frac{d}{dx}(y \cdot \log(\tan x)) = \frac{d}{dx}(x \cdot \log(\tan y)) \]

Using the product rule, we have:
\[ \frac{dy}{dx} \cdot \log(\tan x) + y \cdot \frac{d}{dx}(\log(\tan x)) = \log(\tan y) + x \cdot \frac{d}{dx}(\log(\tan y)) \]

### Step 4: Differentiate the logarithmic terms
Now we differentiate the logarithmic terms:
\[ \frac{d}{dx}(\log(\tan x)) = \frac{1}{\tan x} \cdot \sec^2 x \]
\[ \frac{d}{dx}(\log(\tan y)) = \frac{1}{\tan y} \cdot \sec^2 y \cdot \frac{dy}{dx} \]

Substituting these derivatives back into our equation gives:
\[ \frac{dy}{dx} \cdot \log(\tan x) + y \cdot \left(\frac{\sec^2 x}{\tan x}\right) = \log(\tan y) + x \cdot \left(\frac{\sec^2 y}{\tan y} \cdot \frac{dy}{dx}\right) \]

### Step 5: Rearranging the equation
Now we rearrange the equation to isolate \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} \cdot \log(\tan x) - x \cdot \left(\frac{\sec^2 y}{\tan y} \cdot \frac{dy}{dx}\right) = \log(\tan y) - y \cdot \left(\frac{\sec^2 x}{\tan x}\right) \]

Factoring out \( \frac{dy}{dx} \) gives:
\[ \frac{dy}{dx} \left(\log(\tan x) - x \cdot \frac{\sec^2 y}{\tan y}\right) = \log(\tan y) - y \cdot \frac{\sec^2 x}{\tan x} \]

### Step 6: Solve for \( \frac{dy}{dx} \)
Finally, we solve for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{\log(\tan y) - y \cdot \frac{\sec^2 x}{\tan x}}{\log(\tan x) - x \cdot \frac{\sec^2 y}{\tan y}} \]

### Final Answer
Thus, the derivative \( \frac{dy}{dx} \) is given by:
\[ \frac{dy}{dx} = \frac{\log(\tan y) - y \cdot \frac{\sec^2 x}{\tan x}}{\log(\tan x) - x \cdot \frac{\sec^2 y}{\tan y}} \]