The function provided is:

$f(x) = \frac{e^x + e^{-x}}{2}$

To find the derivative $f'(x)$, we can differentiate this expression.

Step 1: Rewrite the function

$f(x) = \frac{1}{2}(e^x + e^{-x})$ This allows us to differentiate each term more easily.

Step 2: Differentiate each term

Using basic differentiation rules:

• The derivative of $e^x$ is $e^x$.
• The derivative of $e^{-x}$ is $-e^{-x}$, applying the chain rule.

Thus, the derivative of $f(x)$ is: $f'(x) = \frac{1}{2} \left( e^x - e^{-x} \right)$

This simplifies to: $f'(x) = \frac{e^x - e^{-x}}{2}$

Final Answer:

$f'(x) = \frac{e^x - e^{-x}}{2}$

This is the derivative of the given function.

Let me know if you'd like more details or have any questions!

Here are 5 related questions you might find helpful:

1. What is the interpretation of this derivative function geometrically?
2. How can we verify that this function is even or odd based on the derivative?
3. What is the second derivative of the function $f(x)$?
4. Can you express $f(x)$ in terms of hyperbolic functions?
5. How does this function behave as $x \to \infty$ or $x \to -\infty$?

Tip: This function is related to hyperbolic cosine $\cosh(x)$, as $f(x) = \cosh(x)$. You can express the derivative in terms of hyperbolic sine $\sinh(x)$.