To find the derivative of the function $f(x) = x \ln x - x$, we will use the rules of differentiation.

Function:

$f(x) = x \ln x - x$

Step 1: Differentiate $x \ln x$

We apply the product rule, which states that if $u(x)$ and $v(x)$ are functions of $x$, then the derivative of their product is: $\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$

Here, $u(x) = x$ and $v(x) = \ln x$.

• The derivative of $u(x) = x$ is $1$.
• The derivative of $v(x) = \ln x$ is $\frac{1}{x}$.

Now apply the product rule: $\frac{d}{dx}(x \ln x) = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1$

Step 2: Differentiate $-x$

The derivative of $-x$ is simply $-1$.

Step 3: Combine the results

Now, combining the two results: $f'(x) = \ln x + 1 - 1 = \ln x$

Final Answer:

$f'(x) = \ln x$

Would you like further details or have any questions?

Related Questions:

1. What are the critical points of $f(x) = x \ln x - x$?
2. What is the second derivative of $f(x) = x \ln x - x$?
3. How does the function $f(x) = x \ln x - x$ behave at $x = 1$?
4. What is the domain of the function $f(x) = x \ln x - x$?
5. How can the derivative $f'(x) = \ln x$ be used to determine concavity?

Tip:

Always remember to check the domain of the logarithmic function when differentiating expressions involving $\ln x$, since $\ln x$ is only defined for $x > 0$.