Differentiate the following w.r.t. x :
`(x)^(tanx)+(tanx)^(x)`

Answer

To differentiate the function \( y = x^{\tan x} + (\tan x)^x \) with respect to \( x \), we will use the properties of logarithms and differentiation rules such as the product rule and chain rule.

### Step-by-Step Solution

1. **Define the Function:**
\[ y = x^{\tan x} + (\tan x)^x \]

2. **Differentiate Each Term:**
We will differentiate \( y \) term by term. Let:
\[ u = x^{\tan x} \quad \text{and} \quad v = (\tan x)^x \]
Then, we have:
\[ \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \]

3. **Differentiate \( u = x^{\tan x} \):**
To differentiate \( u \), we take the natural logarithm:
\[ \ln u = \tan x \cdot \ln x \]
Now, differentiate both sides:
\[ \frac{1}{u} \frac{du}{dx} = \sec^2 x \cdot \ln x + \tan x \cdot \frac{1}{x} \]
Rearranging gives:
\[ \frac{du}{dx} = u \left( \sec^2 x \cdot \ln x + \tan x \cdot \frac{1}{x} \right) \]
Substituting back for \( u \):
\[ \frac{du}{dx} = x^{\tan x} \left( \sec^2 x \cdot \ln x + \tan x \cdot \frac{1}{x} \right) \]

4. **Differentiate \( v = (\tan x)^x \):**
Again, take the natural logarithm:
\[ \ln v = x \cdot \ln(\tan x) \]
Differentiate:
\[ \frac{1}{v} \frac{dv}{dx} = \ln(\tan x) + x \cdot \frac{1}{\tan x} \cdot \sec^2 x \]
Rearranging gives:
\[ \frac{dv}{dx} = v \left( \ln(\tan x) + x \cdot \frac{\sec^2 x}{\tan x} \right) \]
Substituting back for \( v \):
\[ \frac{dv}{dx} = (\tan x)^x \left( \ln(\tan x) + x \cdot \frac{\sec^2 x}{\tan x} \right) \]

5. **Combine the Derivatives:**
Now we can combine the derivatives:
\[ \frac{dy}{dx} = x^{\tan x} \left( \sec^2 x \cdot \ln x + \tan x \cdot \frac{1}{x} \right) + (\tan x)^x \left( \ln(\tan x) + x \cdot \frac{\sec^2 x}{\tan x} \right) \]

### Final Answer:
\[ \frac{dy}{dx} = x^{\tan x} \left( \sec^2 x \cdot \ln x + \frac{\tan x}{x} \right) + (\tan x)^x \left( \ln(\tan x) + x \cdot \frac{\sec^2 x}{\tan x} \right) \]