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To find the derivative of the function \( f(t) = \frac{\cos(t)}{t^5} \) using the quotient rule, let's follow the steps carefully.

### The quotient rule states:
\[
\frac{d}{dt} \left( \frac{u(t)}{v(t)} \right) = \frac{v(t) \cdot u'(t) - u(t) \cdot v'(t)}{v(t)^2}
\]
Where:
- \( u(t) = \cos(t) \)
- \( v(t) = t^5 \)

### Step 1: Find the derivatives of \( u(t) \) and \( v(t) \)
- \( u'(t) = \frac{d}{dt}[\cos(t)] = -\sin(t) \)
- \( v'(t) = \frac{d}{dt}[t^5] = 5t^4 \)

### Step 2: Apply the quotient rule
\[
f'(t) = \frac{t^5 \cdot (-\sin(t)) - \cos(t) \cdot 5t^4}{(t^5)^2}
\]
Simplify the numerator:
\[
f'(t) = \frac{-t^5 \sin(t) - 5t^4 \cos(t)}{t^{10}}
\]
### Step 3: Simplify the entire expression
Now, simplify each term:
\[
f'(t) = \frac{-t^5 \sin(t)}{t^{10}} - \frac{5t^4 \cos(t)}{t^{10}}
\]
This simplifies further to:
\[
f'(t) = -\frac{\sin(t)}{t^5} - \frac{5 \cos(t)}{t^6}
\]

Thus, the derivative is:
\[
f'(t) = -\frac{\sin(t)}{t^5} - \frac{5 \cos(t)}{t^6}
\]

Would you like more details or have any questions about this solution?

Here are some related questions you might want to explore:
1. How is the quotient rule derived?
2. What happens if \( t = 0 \) in this function? (Considering undefined behavior)
3. Can this function be differentiated using another method?
4. How does the derivative behave as \( t \to \infty \)?
5. What is the practical significance of this derivative in physics?

**Tip:** Always remember to check if the function is defined for all values in its domain, especially when using the quotient rule.

Use the Quotient Rule to find the derivative of the function f(t) = cos(t) / t^5.

Learn how to differentiate the function f(t) = cos(t) / t^5 using the quotient rule. This problem demonstrates key calculus concepts, such as applying the quotient rule and simplifying derivatives involving trigonometric and polynomial functions. Suitable for students in Grades 11-12 or early college-level calculus.

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