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

Answer

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

### Step 1: Take the natural logarithm of both sides
We start by taking the natural logarithm of both sides to simplify the differentiation:
\[ \ln y = \ln(x^{\sqrt{x}}) \]

### Step 2: Apply the logarithmic identity
Using the property of logarithms that states \( \ln(a^b) = b \ln a \), we can rewrite the right side:
\[ \ln y = \sqrt{x} \cdot \ln x \]

### Step 3: Differentiate both sides with respect to \( x \)
Now we differentiate both sides. The left side requires the chain rule, while the right side requires the product rule:
\[ \frac{d}{dx}(\ln y) = \frac{d}{dx}(\sqrt{x} \ln x) \]
Using the chain rule on the left side:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(\sqrt{x}) \cdot \ln x + \sqrt{x} \cdot \frac{d}{dx}(\ln x) \]

### Step 4: Differentiate the right side
Now we differentiate the right side:
- The derivative of \( \sqrt{x} \) is \( \frac{1}{2\sqrt{x}} \).
- The derivative of \( \ln x \) is \( \frac{1}{x} \).

So we have:
\[ \frac{1}{y} \frac{dy}{dx} = \left(\frac{1}{2\sqrt{x}} \cdot \ln x\right) + \left(\sqrt{x} \cdot \frac{1}{x}\right) \]
This simplifies to:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{\ln x}{2\sqrt{x}} + \frac{1}{\sqrt{x}} \]

### Step 5: Solve for \( \frac{dy}{dx} \)
Now we can solve for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = y \left(\frac{\ln x}{2\sqrt{x}} + \frac{1}{\sqrt{x}}\right) \]

### Step 6: Substitute back for \( y \)
Since \( y = x^{\sqrt{x}} \), we substitute back:
\[ \frac{dy}{dx} = x^{\sqrt{x}} \left(\frac{\ln x}{2\sqrt{x}} + \frac{1}{\sqrt{x}}\right) \]

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

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