Differentiate the following w.r.t.x.
`x^(x)sqrtx`

Answer

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

### Step 1: Rewrite the function
First, we rewrite the function for clarity:
\[ y = x^x \cdot x^{1/2} \]
This simplifies to:
\[ y = x^{x + 1/2} \]

### Step 2: Take the natural logarithm
Next, we take the natural logarithm of both sides to facilitate differentiation:
\[ \ln y = \ln(x^{x + 1/2}) \]
Using the property of logarithms, this becomes:
\[ \ln y = (x + \frac{1}{2}) \ln x \]

### Step 3: Differentiate both sides
Now we differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(\ln y) = \frac{d}{dx}((x + \frac{1}{2}) \ln x) \]
Using the chain rule on the left side:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(x \ln x + \frac{1}{2} \ln x) \]

### Step 4: Differentiate the right side
Now we differentiate the right side using the product rule:
1. For \( x \ln x \):
\[ \frac{d}{dx}(x \ln x) = \ln x + 1 \]
2. For \( \frac{1}{2} \ln x \):
\[ \frac{d}{dx}(\frac{1}{2} \ln x) = \frac{1}{2} \cdot \frac{1}{x} \]

Combining these results gives:
\[ \frac{d}{dx}((x + \frac{1}{2}) \ln x) = \ln x + 1 + \frac{1}{2x} \]

### Step 5: Substitute back
Now we substitute back into our equation:
\[ \frac{1}{y} \frac{dy}{dx} = \ln x + 1 + \frac{1}{2x} \]

### Step 6: Solve for \( \frac{dy}{dx} \)
Multiplying both sides by \( y \):
\[ \frac{dy}{dx} = y \left( \ln x + 1 + \frac{1}{2x} \right) \]
Substituting \( y = x^{x + 1/2} \):
\[ \frac{dy}{dx} = x^{x + 1/2} \left( \ln x + 1 + \frac{1}{2x} \right) \]

### Final Answer
Thus, the derivative of \( y = x^x \sqrt{x} \) with respect to \( x \) is:
\[ \frac{dy}{dx} = x^{x + 1/2} \left( \ln x + 1 + \frac{1}{2x} \right) \]

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