Find the derivative of the function.
​m(t)equalsnegative 6 t left parenthesis 5 t Superscript 8 Baseline minus 1 right parenthesis Superscript 8
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Part 1
m prime left parenthesis t right parenthesisequals
  
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To find the derivative of the function \( m(t) = -6t (5t^8 - 1)^8 \), we can use the product rule and the chain rule.

### Step 1: Recognize the Product Rule
The function \( m(t) \) is a product of two parts:
1. \( -6t \)
2. \( (5t^8 - 1)^8 \)

The product rule states:
\[
(m \cdot n)' = m' \cdot n + m \cdot n'
\]
where \( m = -6t \) and \( n = (5t^8 - 1)^8 \).

### Step 2: Differentiate Each Part

1. **Differentiate \( m = -6t \):**
   \[
   m' = -6
   \]

2. **Differentiate \( n = (5t^8 - 1)^8 \):**
   Here we use the chain rule. Let \( u = 5t^8 - 1 \), so \( n = u^8 \).
   - Differentiate \( u^8 \) with respect to \( u \): \( \frac{d}{du}(u^8) = 8u^7 \).
   - Then, differentiate \( u = 5t^8 - 1 \) with respect to \( t \): \( \frac{d}{dt}(5t^8 - 1) = 40t^7 \).

   Combining these using the chain rule:
   \[
   n' = 8(5t^8 - 1)^7 \cdot 40t^7 = 320t^7(5t^8 - 1)^7
   \]

### Step 3: Apply the Product Rule
Now, we can find \( m'(t) \):
\[
m'(t) = m' \cdot n + m \cdot n'
\]
Substituting \( m' = -6 \), \( n = (5t^8 - 1)^8 \), \( m = -6t \), and \( n' = 320t^7(5t^8 - 1)^7 \):

\[
m'(t) = (-6)(5t^8 - 1)^8 + (-6t)(320t^7(5t^8 - 1)^7)
\]

### Step 4: Simplify
Simplify each term:
1. The first term is:
   \[
   -6(5t^8 - 1)^8
   \]

2. The second term is:
   \[
   -6t \cdot 320t^7(5t^8 - 1)^7 = -1920t^8(5t^8 - 1)^7
   \]

So, combining both terms:
\[
m'(t) = -6(5t^8 - 1)^8 - 1920t^8(5t^8 - 1)^7
\]

This is the derivative of \( m(t) \). Let me know if you'd like more details on any step! 

### Additional Questions for Practice
1. How would the derivative change if the exponent in the second part was different?
2. What if the function included an additional trigonometric factor, like \( \sin(t) \)?
3. How would we apply the chain rule if we had \( m(t) = -6t \sin((5t^8 - 1)^8) \)?
4. What if the outermost exponent was a variable instead of 8, such as \( m(t) = -6t (5t^8 - 1)^x \)?
5. Can you find the second derivative, \( m''(t) \), for this function?

**Tip:** Always remember to check if you can simplify each component of a derivative before combining them!

Find the derivative of the function m(t) = -6t(5t^8 - 1)^8.

Explore the step-by-step solution for finding the derivative of m(t) = -6t(5t^8 - 1)^8. This problem demonstrates the use of the product and chain rules in calculus to differentiate complex functions. Suitable for students in Grade 12 or college calculus courses.

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