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We are tasked with finding the derivative of the function:

\[
f(x) = x \sinh(x) - 5 \cosh(x)
\]

Let's compute \( f'(x) \):

### Step 1: Derivative of \( x \sinh(x) \)
Using the product rule: \( \frac{d}{dx}(u \cdot v) = u'v + uv' \), where \( u = x \) and \( v = \sinh(x) \).
- \( u' = 1 \)
- \( v' = \cosh(x) \)

Thus, the derivative of \( x \sinh(x) \) is:
\[
\frac{d}{dx}[x \sinh(x)] = 1 \cdot \sinh(x) + x \cdot \cosh(x) = \sinh(x) + x \cosh(x)
\]

### Step 2: Derivative of \( -5 \cosh(x) \)
The derivative of \( \cosh(x) \) is \( \sinh(x) \). So,
\[
\frac{d}{dx}[-5 \cosh(x)] = -5 \sinh(x)
\]

### Step 3: Combine the results
Now, combining the derivatives from Step 1 and Step 2, we get:
\[
f'(x) = \sinh(x) + x \cosh(x) - 5 \sinh(x)
\]

Simplifying this:
\[
f'(x) = (1 - 5)\sinh(x) + x \cosh(x) = -4 \sinh(x) + x \cosh(x)
\]

Thus, the derivative is:

\[
f'(x) = -4 \sinh(x) + x \cosh(x)
\]

Would you like further details or explanations on any step?

Here are 5 related questions:
1. What is the product rule, and how does it apply to functions like \( x \sinh(x) \)?
2. What are the derivatives of hyperbolic functions like \( \sinh(x) \) and \( \cosh(x) \)?
3. How can you simplify expressions involving hyperbolic functions?
4. What other properties do hyperbolic functions share with trigonometric functions?
5. Can this derivative be used to find critical points of the function?

**Tip:** When working with hyperbolic functions, their derivatives closely resemble those of trigonometric functions, so they are easy to remember!

Find the derivative of f(x) = x sinh(x) - 5 cosh(x).

This problem involves finding the derivative of the function f(x) = x sinh(x) - 5 cosh(x) using the product rule. The solution includes the derivative of hyperbolic functions and their properties, suitable for students studying calculus at the high school level.

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