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

Answer

To differentiate the function \( y = (1+x)^{\log x} \) with respect to \( x \), we can follow these steps:

### Step 1: Take the logarithm of both sides
We start by taking the natural logarithm of both sides to simplify the differentiation process:

\[ \log y = \log((1+x)^{\log x}) \]

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

\[ \log y = \log x \cdot \log(1+x) \]

### Step 3: Differentiate both sides
Now, we differentiate both sides with respect to \( x \). We will use the product rule on the right-hand side:

\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\log x) \cdot \log(1+x) + \log x \cdot \frac{d}{dx}(\log(1+x)) \]

### Step 4: Differentiate the components
Now we differentiate each part:
- The derivative of \( \log x \) is \( \frac{1}{x} \).
- The derivative of \( \log(1+x) \) is \( \frac{1}{1+x} \).

Substituting these derivatives back into our equation gives:

\[ \frac{1}{y} \frac{dy}{dx} = \frac{1}{x} \log(1+x) + \log x \cdot \frac{1}{1+x} \]

### Step 5: Solve for \( \frac{dy}{dx} \)
Now, we can multiply both sides by \( y \) to isolate \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = y \left( \frac{1}{x} \log(1+x) + \frac{\log x}{1+x} \right) \]

### Step 6: Substitute back for \( y \)
Recall that \( y = (1+x)^{\log x} \). Thus, we substitute \( y \) back into the equation:

\[ \frac{dy}{dx} = (1+x)^{\log x} \left( \frac{1}{x} \log(1+x) + \frac{\log x}{1+x} \right) \]

### Final Result
The derivative of \( y = (1+x)^{\log x} \) with respect to \( x \) is:

\[ \frac{dy}{dx} = (1+x)^{\log x} \left( \frac{1}{x} \log(1+x) + \frac{\log x}{1+x} \right) \]

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