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

Options

A. `(log(tany))/(lot(tanx))`
B. `(lot(tany)-1)/(2log(tanx))`
C. `lot(tany)-(2y)/(2sin2x)`
D. `(lot(tany)-2ycosec2x)/(log(tanx)-2x cosec 2y)`

Answer

To solve the equation \((\tan x)^y = (\tan y)^x\) 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:

\[ \log((\tan x)^y) = \log((\tan y)^x) \]

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

\[ y \log(\tan x) = x \log(\tan y) \]

### 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.

The left-hand side becomes:

\[ \frac{d}{dx}(y \log(\tan x)) = \frac{dy}{dx} \log(\tan x) + y \cdot \frac{d}{dx}(\log(\tan x)) \]

Using the chain rule, we find:

\[ \frac{d}{dx}(\log(\tan x)) = \frac{1}{\tan x} \cdot \sec^2 x = \frac{\sec^2 x}{\tan x} \]

So, the left-hand side becomes:

\[ \frac{dy}{dx} \log(\tan x) + y \cdot \frac{\sec^2 x}{\tan x} \]

The right-hand side becomes:

\[ \frac{d}{dx}(x \log(\tan y)) = \log(\tan y) + x \cdot \frac{d}{dx}(\log(\tan y)) \]

Using the chain rule again, we have:

\[ \frac{d}{dx}(\log(\tan y)) = \frac{1}{\tan y} \cdot \sec^2 y \cdot \frac{dy}{dx} \]

Thus, the right-hand side becomes:

\[ \log(\tan y) + x \cdot \frac{\sec^2 y}{\tan y} \cdot \frac{dy}{dx} \]

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

\[ \frac{dy}{dx} \log(\tan x) + y \cdot \frac{\sec^2 x}{\tan x} = \log(\tan y) + x \cdot \frac{\sec^2 y}{\tan y} \cdot \frac{dy}{dx} \]

### Step 5: Collect \(\frac{dy}{dx}\) terms
Rearranging gives:

\[ \frac{dy}{dx} \log(\tan x) - x \cdot \frac{\sec^2 y}{\tan y} \cdot \frac{dy}{dx} = \log(\tan y) - y \cdot \frac{\sec^2 x}{\tan x} \]

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

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

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

\[ \frac{dy}{dx} = \frac{\log(\tan y) - y \cdot \frac{\sec^2 x}{\tan x}}{\log(\tan x) - x \cdot \frac{\sec^2 y}{\tan y}} \]

### Summary of the solution
Thus, the derivative \(\frac{dy}{dx}\) is given by:

\[ \frac{dy}{dx} = \frac{\log(\tan y) - y \cdot \frac{\sec^2 x}{\tan x}}{\log(\tan x) - x \cdot \frac{\sec^2 y}{\tan y}} \]