Compute f '(a) algebraically for the given value of a. HINT [See Example 1.]
f(x) = x3 + 5x;  a = 6

To compute the derivative \( f'(a) \) for the function \( f(x) = x^3 + 5x \) at \( a = 6 \), we can follow these steps:

### Step 1: Differentiate the function \( f(x) \)

To find \( f'(x) \), we differentiate the given function:

\[
f(x) = x^3 + 5x
\]

Using basic differentiation rules:
- The derivative of \( x^3 \) is \( 3x^2 \).
- The derivative of \( 5x \) is \( 5 \).

So, the derivative of \( f(x) \) is:

\[
f'(x) = 3x^2 + 5
\]

### Step 2: Evaluate the derivative at \( a = 6 \)

Now, we substitute \( x = 6 \) into the derivative:

\[
f'(6) = 3(6)^2 + 5
\]

First, compute \( 6^2 \):

\[
6^2 = 36
\]

Now substitute it into the equation:

\[
f'(6) = 3(36) + 5 = 108 + 5 = 113
\]

Thus, the value of \( f'(6) \) is:

\[
f'(6) = 113
\]

### Final Answer:

\[
f'(6) = 113
\]

Would you like more details or have any questions?

### Relative Questions:
1. How do you differentiate \( x^n \) in general?
2. Why is the derivative of \( 5x \) simply \( 5 \)?
3. Can you explain how to apply the power rule in calculus?
4. What happens to the derivative when constants are added to a function?
5. What is the meaning of \( f'(a) \) in terms of slope at a point?

### Tip:
When differentiating, remember that constant terms (like 5 in this case) remain unchanged when multiplied with variables, but their derivatives themselves depend on the degree of the variable.

Compute f '(a) algebraically for the given value of a. HINT [See Example 1.] f(x) = x^3 + 5x;  a = 6

Learn how to compute the derivative f'(6) for the function f(x) = x^3 + 5x using the power rule. This step-by-step solution covers key differentiation concepts for polynomial functions. Suitable for students in Grades 10-12.

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