To find the derivative $f'(x)$ and evaluate it at $x = 5$, we will first compute the derivative of the function:

$f(x) = \frac{1}{2}x^2 + x^3 - 2x.$

Step 1: Differentiate $f(x)$

Using the rules of differentiation:

1. The derivative of $\frac{1}{2}x^2$ is $\frac{1}{2} \cdot 2x = x$.
2. The derivative of $x^3$ is $3x^2$.
3. The derivative of $-2x$ is $-2$.

Thus, the derivative $f'(x)$ is:

$f'(x) = x + 3x^2 - 2.$

Step 2: Evaluate $f'(5)$

Now, substitute $x = 5$ into $f'(x)$:

$f'(5) = 5 + 3 \cdot 5^2 - 2.$

Calculate the value:

$f'(5) = 5 + 3 \cdot 25 - 2 = 5 + 75 - 2 = 78.$

Final Answer

$f'(5) = 78.$

Would you like further details or explanations? Here are some related questions to explore:

1. What is the second derivative, $f''(x)$, of the function?
2. How do you find the critical points of $f(x)$?
3. Can we determine the concavity of $f(x)$ using the second derivative?
4. What is the value of $f(x)$ at $x = 5$?
5. How would you find the local minima and maxima of $f(x)$?

Tip: When finding derivatives, always remember the power rule: if $f(x) = ax^n$, then $f'(x) = n \cdot ax^{n-1}$.