find the second order derivative of the function $$ f\left(x\right)=\sqrt{x}\ln x $$

We are tasked with finding the second-order derivative of the function $ f(x) = \sqrt{x} \ln(x) $.

### Step 1: Rewrite the function
First, express the function in a more convenient form for differentiation:

\[
f(x) = \sqrt{x} \ln(x) = x^{1/2} \ln(x)
\]

### Step 2: First derivative
To differentiate $ f(x) = x^{1/2} \ln(x) $, we'll use the **product rule**, which states:

\[
\frac{d}{dx} \left( u(x) v(x) \right) = u'(x) v(x) + u(x) v'(x)
\]

In this case:
- $ u(x) = x^{1/2} $, so $ u'(x) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}} $
- $ v(x) = \ln(x) $, so $ v'(x) = \frac{1}{x} $

Now apply the product rule:

\[
f'(x) = u'(x) v(x) + u(x) v'(x)
\]

Substitute the derivatives:

\[
f'(x) = \frac{1}{2\sqrt{x}} \ln(x) + x^{1/2} \cdot \frac{1}{x}
\]

Simplify the second term:

\[
f'(x) = \frac{1}{2\sqrt{x}} \ln(x) + \frac{1}{\sqrt{x}}
\]

So the first derivative is:

\[
f'(x) = \frac{\ln(x)}{2\sqrt{x}} + \frac{1}{\sqrt{x}}
\]

### Step 3: Second derivative
Now we'll differentiate $ f'(x) = \frac{\ln(x)}{2\sqrt{x}} + \frac{1}{\sqrt{x}} $.

We'll differentiate each term separately.

#### First term: $ \frac{\ln(x)}{2\sqrt{x}} $

We apply the **quotient rule** here:

\[
\frac{d}{dx} \left( \frac{\ln(x)}{2\sqrt{x}} \right) = \frac{\left( \frac{1}{x} \right)(2\sqrt{x}) - (\ln(x)) \left( \frac{d}{dx}(2\sqrt{x}) \right)}{(2\sqrt{x})^2}
\]

We already know $ \frac{d}{dx}(2\sqrt{x}) = \frac{1}{\sqrt{x}} $. Now simplify:

\[
= \frac{\frac{1}{x} \cdot 2\sqrt{x} - \ln(x) \cdot \frac{1}{\sqrt{x}}}{4x}
\]

\[
= \frac{2 \cdot \frac{1}{\sqrt{x}} - \frac{\ln(x)}{\sqrt{x}}}{4x}
\]

\[
= \frac{(2 - \ln(x))}{4x\sqrt{x}}
\]

#### Second term: $ \frac{1}{\sqrt{x}} $

The derivative of $ \frac{1}{\sqrt{x}} = x^{-1/2} $ is:

\[
\frac{d}{dx}\left(x^{-1/2}\right) = -\frac{1}{2}x^{-3/2} = -\frac{1}{2x\sqrt{x}}
\]

### Step 4: Combine the results

Now combine both terms to get the second derivative:

\[
f''(x) = \frac{2 - \ln(x)}{4x\sqrt{x}} - \frac{1}{2x\sqrt{x}}
\]

We can combine these fractions since they have the same denominator:

\[
f''(x) = \frac{2 - \ln(x) - 2}{4x\sqrt{x}}
\]

\[
f''(x) = \frac{-\ln(x)}{4x\sqrt{x}}
\]

### Final Answer:
The second-order derivative of $ f(x) = \sqrt{x} \ln(x) $ is:

\[
f''(x) = \frac{-\ln(x)}{4x\sqrt{x}}
\]

Would you like further details or have any questions?

Here are some related questions:
1. How would you differentiate $ f(x) = x^{3/2} \ln(x) $?
2. How does the product rule differ from the chain rule?
3. Can you find the second derivative of $ f(x) = x \ln(x) $?
4. How would you integrate $ f'(x) = \frac{\ln(x)}{2\sqrt{x}} + \frac{1}{\sqrt{x}} $?
5. What is the significance of higher-order derivatives in real-world problems?

**Tip:** When working with logarithmic and power functions together, it's helpful to rewrite the expressions in simpler forms to avoid mistakes during differentiation.

Find the second order derivative of the function $$ f(x)=\sqrt{x}\ln x $$

Learn how to find the second-order derivative of the function f(x) = sqrt(x) ln(x). This calculus problem involves using the product and quotient rules, covering logarithmic functions and advanced differentiation techniques. Ideal for college-level calculus students.

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