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

Answer

To differentiate the function \( y = x^{\log x} + (\log x)^{x} \) with respect to \( x \), we will follow these steps:

### Step 1: Rewrite the function
We start by letting:
\[ y = x^{\log x} + (\log x)^{x} \]

### Step 2: Differentiate each term
We will differentiate each term separately.

#### Differentiate \( x^{\log x} \)
Using the property \( a^b = e^{b \ln a} \), we can rewrite \( x^{\log x} \):
\[ x^{\log x} = e^{\log x \cdot \ln x} \]
Now, we differentiate using the chain rule:
\[ \frac{dy_1}{dx} = e^{\log x \cdot \ln x} \cdot \frac{d}{dx}(\log x \cdot \ln x) \]
Using the product rule on \( \log x \cdot \ln x \):
\[ \frac{d}{dx}(\log x \cdot \ln x) = \ln x \cdot \frac{d}{dx}(\log x) + \log x \cdot \frac{d}{dx}(\ln x) \]
Calculating the derivatives:
\[ \frac{d}{dx}(\log x) = \frac{1}{x \ln 10}, \quad \frac{d}{dx}(\ln x) = \frac{1}{x} \]
Thus,
\[ \frac{d}{dx}(\log x \cdot \ln x) = \ln x \cdot \frac{1}{x \ln 10} + \log x \cdot \frac{1}{x} \]

#### Differentiate \( (\log x)^{x} \)
Using the same exponential property:
\[ (\log x)^{x} = e^{x \ln(\log x)} \]
Differentiating:
\[ \frac{dy_2}{dx} = e^{x \ln(\log x)} \cdot \frac{d}{dx}(x \ln(\log x)) \]
Using the product rule:
\[ \frac{d}{dx}(x \ln(\log x)) = \ln(\log x) + x \cdot \frac{1}{\log x} \cdot \frac{1}{x} = \ln(\log x) + \frac{1}{\log x} \]

### Step 3: Combine the derivatives
Now we combine the derivatives:
\[ \frac{dy}{dx} = \frac{dy_1}{dx} + \frac{dy_2}{dx} \]
Substituting back:
\[ \frac{dy}{dx} = x^{\log x} \left( \ln x \cdot \frac{1}{x \ln 10} + \log x \cdot \frac{1}{x} \right) + (\log x)^{x} \left( \ln(\log x) + \frac{1}{\log x} \right) \]

### Final Result
Thus, the derivative of the function is:
\[ \frac{dy}{dx} = x^{\log x} \left( \frac{\ln x}{x \ln 10} + \frac{\log x}{x} \right) + (\log x)^{x} \left( \ln(\log x) + \frac{1}{\log x} \right) \]

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