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To find the derivative of the function:

\[
f(t) = (4 + \sqrt{t})(2t^2 - 7)
\]

we will need to apply the **Product Rule**, since this is a product of two functions. The Product Rule is:

\[
\frac{d}{dt}[u(t)v(t)] = u'(t)v(t) + u(t)v'(t)
\]

Here, \( u(t) = 4 + \sqrt{t} \) and \( v(t) = 2t^2 - 7 \).

### Step-by-step derivative calculation:

1. **Differentiate** \( u(t) = 4 + \sqrt{t} \):
   \[
   u'(t) = \frac{d}{dt}(4 + \sqrt{t}) = \frac{d}{dt}(4) + \frac{d}{dt}(t^{1/2}) = 0 + \frac{1}{2}t^{-1/2}
   \]
   So, \( u'(t) = \frac{1}{2\sqrt{t}} \).

2. **Differentiate** \( v(t) = 2t^2 - 7 \):
   \[
   v'(t) = \frac{d}{dt}(2t^2 - 7) = 4t
   \]

3. **Apply the Product Rule**:
   \[
   f'(t) = u'(t)v(t) + u(t)v'(t)
   \]
   Substituting the expressions for \( u(t) \), \( u'(t) \), \( v(t) \), and \( v'(t) \):
   \[
   f'(t) = \left( \frac{1}{2\sqrt{t}} \right)(2t^2 - 7) + (4 + \sqrt{t})(4t)
   \]

4. **Simplify the terms**:
   First term:
   \[
   \left( \frac{1}{2\sqrt{t}} \right)(2t^2 - 7) = \frac{2t^2 - 7}{2\sqrt{t}}
   \]

   Second term:
   \[
   (4 + \sqrt{t})(4t) = 16t + 4t^{3/2}
   \]

   Therefore, the derivative is:
   \[
   f'(t) = \frac{2t^2 - 7}{2\sqrt{t}} + 16t + 4t^{3/2}
   \]

This is the derivative of the function.

Let me know if you'd like more details on any specific part!

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Here are some related questions for deeper understanding:

1. How would you apply the product rule to other types of functions?
2. What is the importance of simplifying terms in derivative calculations?
3. Could you find the second derivative of \( f(t) \)?
4. How does the chain rule relate to this example, particularly in \( \sqrt{t} \)?
5. How does the product rule differ from the quotient rule?

**Tip**: Always identify the type of operation (product, quotient, chain) in a function before taking the derivative to apply the correct rule!

Find the derivative of the function f(t) = (4 + sqrt(t))(2t^2 - 7)

Learn how to find the derivative of f(t) = (4 + sqrt(t))(2t^2 - 7) using the product rule. This solution explains the product rule and simplifies the expression step-by-step. Suitable for high school or college-level calculus students.

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