Differentiate the following w.r.t. x :
`sqrtxlogx^(2)`

Answer

To differentiate the function \( y = \sqrt{x} \log(x^2) \) with respect to \( x \), we will use the product rule. The product rule states that if you have two functions \( u \) and \( v \), then:

\[ \frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx} \]

### Step 1: Identify \( u \) and \( v \)
Let:
- \( u = \sqrt{x} \)
- \( v = \log(x^2) \)

### Step 2: Differentiate \( u \) and \( v \)
1. Differentiate \( u \):
\[ u = \sqrt{x} = x^{1/2} \]
Using the power rule:
\[ \frac{du}{dx} = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}} \]

2. Differentiate \( v \):
\[ v = \log(x^2) = 2\log(x) \]
Using the derivative of logarithm:
\[ \frac{dv}{dx} = 2 \cdot \frac{1}{x} = \frac{2}{x} \]

### Step 3: Apply the product rule
Now, applying the product rule:
\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]
Substituting the values of \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \):
\[ \frac{dy}{dx} = \sqrt{x} \cdot \frac{2}{x} + 2\log(x) \cdot \frac{1}{2\sqrt{x}} \]

### Step 4: Simplify the expression
1. The first term:
\[ \sqrt{x} \cdot \frac{2}{x} = \frac{2\sqrt{x}}{x} = \frac{2}{\sqrt{x}} \]

2. The second term:
\[ 2\log(x) \cdot \frac{1}{2\sqrt{x}} = \frac{\log(x)}{\sqrt{x}} \]

Combining both terms:
\[ \frac{dy}{dx} = \frac{2}{\sqrt{x}} + \frac{\log(x)}{\sqrt{x}} \]

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