We are asked to find the first, second, and third derivatives of the function $f(x) = x \ln(x)$.

Step 1: First Derivative $f'(x)$

We use the product rule to differentiate $f(x) = x \ln(x)$.

The product rule states that: $\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$ Here, $u(x) = x$ and $v(x) = \ln(x)$.

Now, calculate their derivatives:

• $u'(x) = \frac{d}{dx}[x] = 1$
• $v'(x) = \frac{d}{dx}[\ln(x)] = \frac{1}{x}$

Now, apply the product rule: $f'(x) = 1 \cdot \ln(x) + x \cdot \frac{1}{x} = \ln(x) + 1$

Step 2: Second Derivative $f''(x)$

Now we differentiate $f'(x) = \ln(x) + 1$.

The derivative of $\ln(x)$ is $\frac{1}{x}$, and the derivative of 1 is 0. So: $f''(x) = \frac{1}{x}$

Step 3: Third Derivative $f'''(x)$

Finally, we differentiate $f''(x) = \frac{1}{x}$.

The derivative of $\frac{1}{x}$ is $-\frac{1}{x^2}$. So: $f'''(x) = -\frac{1}{x^2}$

Summary:

$f'(x) = \ln(x) + 1$ $f''(x) = \frac{1}{x}$ $f'''(x) = -\frac{1}{x^2}$

Would you like further details on these steps or have any questions?

Here are 5 related questions you might consider:

1. How does the product rule apply to more complex functions?
2. What are the conditions for $f'(x)$ to exist for logarithmic functions?
3. Can the chain rule be used in differentiating $x \ln(x)$?
4. What happens to higher-order derivatives of $x \ln(x)$?
5. How do the properties of logarithms affect their derivatives?

Tip: When differentiating a product of two functions, always consider applying the product rule first to avoid mistakes!