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.