Differnetiate `x^(x)w.r.t.xlogx.`

Answer

To differentiate \( u = x^x \) with respect to \( v = x \log x \), we will use the formula:

\[ \frac{du}{dv} = \frac{du/dx}{dv/dx} \]

### Step 1: Differentiate \( u = x^x \)

First, we take the natural logarithm of both sides:

\[ \log u = \log(x^x) = x \log x \]

Now, we differentiate both sides with respect to \( x \):

\[ \frac{d}{dx}(\log u) = \frac{d}{dx}(x \log x) \]

Using the chain rule on the left side:

\[ \frac{1}{u} \frac{du}{dx} = \frac{d}{dx}(x \log x) \]

### Step 2: Differentiate \( x \log x \)

Using the product rule on the right side, where \( f(x) = x \) and \( g(x) = \log x \):

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

### Step 3: Substitute back into the equation

Now we have:

\[ \frac{1}{u} \frac{du}{dx} = 1 + \log x \]

Multiplying both sides by \( u \):

\[ \frac{du}{dx} = u(1 + \log x) \]

Since \( u = x^x \):

\[ \frac{du}{dx} = x^x (1 + \log x) \]

### Step 4: Differentiate \( v = x \log x \)

Now we differentiate \( v \):

\[ v = x \log x \]

Using the product rule again:

\[ \frac{dv}{dx} = x \cdot \frac{1}{x} + \log x \cdot 1 = 1 + \log x \]

### Step 5: Substitute into the formula for \( \frac{du}{dv} \)

Now we can substitute \( \frac{du}{dx} \) and \( \frac{dv}{dx} \) into our original formula:

\[ \frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{x^x (1 + \log x)}{1 + \log x} \]

### Step 6: Simplify

The \( (1 + \log x) \) terms cancel out:

\[ \frac{du}{dv} = x^x \]

### Final Answer

Thus, the derivative of \( x^x \) with respect to \( x \log x \) is:

\[ \frac{du}{dv} = x^x \]

---