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

Answer

To find \(\frac{dy}{dx}\) for the equation \((\tan x)^y = (\tan y)^x\), 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:
\[ \ln((\tan x)^y) = \ln((\tan y)^x) \]

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

### 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:
\[ \frac{d}{dx}(y \ln(\tan x)) = \frac{d}{dx}(x \ln(\tan y)) \]

Applying the product rule:
\[ \frac{dy}{dx} \ln(\tan x) + y \frac{d}{dx}(\ln(\tan x)) = \ln(\tan y) + x \frac{d}{dx}(\ln(\tan y)) \]

### Step 4: Differentiate \(\ln(\tan x)\) and \(\ln(\tan y)\)
Using the chain rule, we find:
\[ \frac{d}{dx}(\ln(\tan x)) = \frac{1}{\tan x} \cdot \sec^2 x = \frac{\sec^2 x}{\tan x} \]
And similarly,
\[ \frac{d}{dx}(\ln(\tan y)) = \frac{1}{\tan y} \cdot \frac{dy}{dx} \sec^2 y \]

### Step 5: Substitute the derivatives back into the equation
Substituting these derivatives back into our differentiated equation gives:
\[ \frac{dy}{dx} \ln(\tan x) + y \frac{\sec^2 x}{\tan x} = \ln(\tan y) + x \left(\frac{1}{\tan y} \cdot \frac{dy}{dx} \sec^2 y\right) \]

### Step 6: Rearranging the equation
Now, we will rearrange the equation to isolate \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} \ln(\tan x) - x \frac{\sec^2 y}{\tan y} \frac{dy}{dx} = \ln(\tan y) - y \frac{\sec^2 x}{\tan x} \]

### Step 7: Factor out \(\frac{dy}{dx}\)
Factoring out \(\frac{dy}{dx}\) gives:
\[ \frac{dy}{dx} \left(\ln(\tan x) - x \frac{\sec^2 y}{\tan y}\right) = \ln(\tan y) - y \frac{\sec^2 x}{\tan x} \]

### Step 8: Solve for \(\frac{dy}{dx}\)
Finally, we can solve for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{\ln(\tan y) - y \frac{\sec^2 x}{\tan x}}{\ln(\tan x) - x \frac{\sec^2 y}{\tan y}} \]

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

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