`y=(tanx)^(cotx)+(cotx)^(tanx)`

Answer

To find the derivative \( \frac{dy}{dx} \) for the function \( y = (\tan x)^{\cot x} + (\cot x)^{\tan x} \), we will break it down into two parts and differentiate each part separately.

### Step-by-Step Solution:

1. **Define the components**:
Let
\[ y_1 = (\tan x)^{\cot x} \quad \text{and} \quad y_2 = (\cot x)^{\tan x} \]
Then,
\[ y = y_1 + y_2 \]

2. **Differentiate \( y \)**:
Using the sum rule of differentiation:
\[ \frac{dy}{dx} = \frac{dy_1}{dx} + \frac{dy_2}{dx} \]

3. **Differentiate \( y_1 \)**:
For \( y_1 = (\tan x)^{\cot x} \), take the natural logarithm:
\[ \log y_1 = \cot x \cdot \log(\tan x) \]
Differentiate both sides with respect to \( x \):
\[ \frac{1}{y_1} \frac{dy_1}{dx} = \frac{d}{dx}(\cot x \cdot \log(\tan x)) \]
Using the product rule:
\[ \frac{d}{dx}(\cot x \cdot \log(\tan x)) = \cot x \cdot \frac{d}{dx}(\log(\tan x)) + \log(\tan x) \cdot \frac{d}{dx}(\cot x) \]
Now calculate \( \frac{d}{dx}(\log(\tan x)) \) and \( \frac{d}{dx}(\cot x) \):
\[ \frac{d}{dx}(\log(\tan x)) = \frac{1}{\tan x} \cdot \sec^2 x = \frac{\sec^2 x}{\tan x} \]
\[ \frac{d}{dx}(\cot x) = -\csc^2 x \]
Thus,
\[ \frac{1}{y_1} \frac{dy_1}{dx} = \cot x \cdot \frac{\sec^2 x}{\tan x} - \log(\tan x) \cdot \csc^2 x \]
Rearranging gives:
\[ \frac{dy_1}{dx} = y_1 \left( \cot x \cdot \frac{\sec^2 x}{\tan x} - \log(\tan x) \cdot \csc^2 x \right) \]

4. **Differentiate \( y_2 \)**:
For \( y_2 = (\cot x)^{\tan x} \), similarly take the natural logarithm:
\[ \log y_2 = \tan x \cdot \log(\cot x) \]
Differentiate:
\[ \frac{1}{y_2} \frac{dy_2}{dx} = \frac{d}{dx}(\tan x \cdot \log(\cot x)) \]
Using the product rule:
\[ \frac{d}{dx}(\tan x \cdot \log(\cot x)) = \tan x \cdot \frac{d}{dx}(\log(\cot x)) + \log(\cot x) \cdot \frac{d}{dx}(\tan x) \]
Calculate \( \frac{d}{dx}(\log(\cot x)) \):
\[ \frac{d}{dx}(\log(\cot x)) = -\frac{1}{\cot x} \cdot \csc^2 x = -\frac{\csc^2 x}{\cot x} \]
Thus,
\[ \frac{1}{y_2} \frac{dy_2}{dx} = \tan x \left(-\frac{\csc^2 x}{\cot x}\right) + \log(\cot x) \sec^2 x \]
Rearranging gives:
\[ \frac{dy_2}{dx} = y_2 \left( -\tan x \cdot \frac{\csc^2 x}{\cot x} + \log(\cot x) \sec^2 x \right) \]

5. **Combine the derivatives**:
Finally, we combine \( \frac{dy_1}{dx} \) and \( \frac{dy_2}{dx} \):
\[ \frac{dy}{dx} = \frac{dy_1}{dx} + \frac{dy_2}{dx} \]

### Final Answer:
The derivative \( \frac{dy}{dx} \) is given by:
\[ \frac{dy}{dx} = \left( (\tan x)^{\cot x} \left( \cot x \cdot \frac{\sec^2 x}{\tan x} - \log(\tan x) \cdot \csc^2 x \right) \right) + \left( (\cot x)^{\tan x} \left( -\tan x \cdot \frac{\csc^2 x}{\cot x} + \log(\cot x) \sec^2 x \right) \right) \]