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

Answer

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

### Step 1: Rewrite the function
First, we rewrite the function using properties of exponents:
\[ y = (\sqrt{x})^{\sqrt{x}} = x^{\frac{1}{2} \cdot \sqrt{x}} = x^{\frac{\sqrt{x}}{2}} \]

### Step 2: Take the natural logarithm of both sides
Taking the natural logarithm of both sides helps us simplify the differentiation:
\[ \ln y = \ln\left(x^{\frac{\sqrt{x}}{2}}\right) \]

### Step 3: Use the logarithmic identity
Using the property of logarithms, we can bring down the exponent:
\[ \ln y = \frac{\sqrt{x}}{2} \ln x \]

### Step 4: Differentiate both sides
Now we differentiate both sides with respect to \( x \). We will use implicit differentiation on the left side and the product rule on the right side:
\[ \frac{d}{dx}(\ln y) = \frac{1}{y} \frac{dy}{dx} \]
For the right side, we apply the product rule:
\[ \frac{d}{dx}\left(\frac{\sqrt{x}}{2} \ln x\right) = \frac{1}{2} \cdot \frac{1}{2\sqrt{x}} \ln x + \frac{\sqrt{x}}{2} \cdot \frac{1}{x} \]
This simplifies to:
\[ \frac{1}{4\sqrt{x}} \ln x + \frac{1}{2\sqrt{x}} \]

### Step 5: Set the derivatives equal
Now we set the derivatives equal to each other:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{1}{4\sqrt{x}} \ln x + \frac{1}{2\sqrt{x}} \]

### Step 6: Solve for \(\frac{dy}{dx}\)
To isolate \(\frac{dy}{dx}\), we multiply both sides by \( y \):
\[ \frac{dy}{dx} = y \left( \frac{1}{4\sqrt{x}} \ln x + \frac{1}{2\sqrt{x}} \right) \]

### Step 7: Substitute back for \( y \)
Recall that \( y = (\sqrt{x})^{\sqrt{x}} \):
\[ \frac{dy}{dx} = (\sqrt{x})^{\sqrt{x}} \left( \frac{1}{4\sqrt{x}} \ln x + \frac{1}{2\sqrt{x}} \right) \]

### Step 8: Simplify the expression
We can factor out \( \frac{1}{\sqrt{x}} \):
\[ \frac{dy}{dx} = \frac{(\sqrt{x})^{\sqrt{x}}}{\sqrt{x}} \left( \frac{1}{4} \ln x + \frac{1}{2} \right) \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{(\sqrt{x})^{\sqrt{x}}}{\sqrt{x}} \left( \frac{1}{4} \ln x + \frac{1}{2} \right) = \frac{(\sqrt{x})^{\sqrt{x} - 1}}{4} \ln x + \frac{(\sqrt{x})^{\sqrt{x}}}{2} \]

### Final Answer
Thus, the derivative of \( y = (\sqrt{x})^{\sqrt{x}} \) with respect to \( x \) is:
\[ \frac{dy}{dx} = \frac{(\sqrt{x})^{\sqrt{x}}}{2} + \frac{(\sqrt{x})^{\sqrt{x} - 1}}{4} \ln x \]