If ` y = x^sqrt(x) , "then "(dy)/(dx)` is equal to

Options

A. `(y)/(sqrt(x)) (2 + log x)`
B. `(y)/(2 sqrt(x)) (2 + log x)`
C. `(y)/(sqrt(x)) (2 - log x)`
D. `(y)/(2sqrt(x)) (2 - log x)`

Explanation

To solve the problem where \( y = x^{\sqrt{x}} \) and we need to find \( \frac{dy}{dx} \), we can follow these steps:

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

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

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

### Step 4: Apply the product rule
The right side involves the product of \( \sqrt{x} \) and \( \ln x \). Using the product rule \( (u \cdot v)' = u'v + uv' \):
- Let \( u = \sqrt{x} \) and \( v = \ln x \).
- Then \( u' = \frac{1}{2\sqrt{x}} \) and \( v' = \frac{1}{x} \).

Applying the product rule:
\[ \frac{d}{dx}(\sqrt{x} \ln x) = \left(\frac{1}{2\sqrt{x}} \cdot \ln x\right) + \left(\sqrt{x} \cdot \frac{1}{x}\right) \]
This simplifies to:
\[ \frac{1}{2\sqrt{x}} \ln x + \frac{1}{\sqrt{x}} \]

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

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

### Step 7: Substitute \( y \) back
Since \( y = x^{\sqrt{x}} \), we substitute back:
\[ \frac{dy}{dx} = x^{\sqrt{x}} \left( \frac{1}{2\sqrt{x}} \ln x + \frac{1}{\sqrt{x}} \right) \]

### Step 8: Combine terms
Factoring out \( \frac{1}{\sqrt{x}} \):
\[ \frac{dy}{dx} = x^{\sqrt{x}} \cdot \frac{1}{\sqrt{x}} \left( \frac{1}{2} \ln x + 1 \right) \]

### Final Result
Thus, the final expression for \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = \frac{x^{\sqrt{x}}}{\sqrt{x}} \left( \frac{1}{2} \ln x + 1 \right) \]