Derivative of ` x ^(x)` w.r.t `xlogx ` is

Options

A. `1+log x`
B. ` x^(x) +log x `
C. ` x^(x) `
D. ` x^(x) -log x `

Answer

To find the derivative of \( x^x \) with respect to \( x \log x \), we will follow these steps:

### Step 1: Define the Functions
Let:
- \( y_1 = x^x \)
- \( y_2 = x \log x \)

We need to find \( \frac{dy_1}{dy_2} \).

### Step 2: Differentiate \( y_1 \)
To differentiate \( y_1 = x^x \), we can use logarithmic differentiation.

Taking the natural logarithm of both sides:
\[ \log y_1 = \log(x^x) = x \log x \]

Now, differentiate both sides with respect to \( x \):
\[ \frac{1}{y_1} \frac{dy_1}{dx} = \log x + x \cdot \frac{1}{x} = \log x + 1 \]

Multiplying both sides by \( y_1 \):
\[ \frac{dy_1}{dx} = y_1 (\log x + 1) = x^x (\log x + 1) \]

### Step 3: Differentiate \( y_2 \)
Now, differentiate \( y_2 = x \log x \):
\[ \frac{dy_2}{dx} = \log x + x \cdot \frac{1}{x} = \log x + 1 \]

### Step 4: Find \( \frac{dy_1}{dy_2} \)
Now, we can find \( \frac{dy_1}{dy_2} \) using the chain rule:
\[ \frac{dy_1}{dy_2} = \frac{dy_1/dx}{dy_2/dx} = \frac{x^x (\log x + 1)}{\log x + 1} \]

### Step 5: Simplify
The \( \log x + 1 \) terms cancel out (assuming \( \log x + 1 \neq 0 \)):
\[ \frac{dy_1}{dy_2} = x^x \]

### Final Answer
Thus, the derivative of \( x^x \) with respect to \( x \log x \) is:
\[ \frac{dy_1}{dy_2} = x^x \]

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