Math Problem Statement

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

Solution

To differentiate the function f(t)=cos(t)t5f(t) = \frac{\cos(t)}{t^5} using the Quotient Rule, recall the rule:

ddt(u(t)v(t))=v(t)u(t)u(t)v(t)v(t)2\frac{d}{dt} \left( \frac{u(t)}{v(t)} \right) = \frac{v(t) \cdot u'(t) - u(t) \cdot v'(t)}{v(t)^2}

Here, u(t)=cos(t)u(t) = \cos(t) and v(t)=t5v(t) = t^5. Now, differentiate u(t)u(t) and v(t)v(t):

  • u(t)=sin(t)u'(t) = -\sin(t)
  • v(t)=5t4v'(t) = 5t^4

Now, apply the Quotient Rule:

f(t)=t5(sin(t))cos(t)(5t4)(t5)2f'(t) = \frac{t^5 \cdot (-\sin(t)) - \cos(t) \cdot (5t^4)}{(t^5)^2}

Simplify the numerator and denominator:

f(t)=t5sin(t)5t4cos(t)t10f'(t) = \frac{-t^5 \sin(t) - 5t^4 \cos(t)}{t^{10}}

Factor the numerator:

f(t)=t4(tsin(t)+5cos(t))t10f'(t) = \frac{-t^4 (t \sin(t) + 5 \cos(t))}{t^{10}}

Finally, simplify:

f(t)=(tsin(t)+5cos(t))t6f'(t) = \frac{-(t \sin(t) + 5 \cos(t))}{t^6}

So, the derivative is:

f(t)=(tsin(t)+5cos(t))t6f'(t) = \frac{-(t \sin(t) + 5 \cos(t))}{t^6}

Would you like further clarification or details on any step? Here are 5 related questions:

  1. What are the steps for applying the Product Rule in a similar function?
  2. How does the Power Rule apply to the differentiation of t5t^5?
  3. What is the significance of factoring in the numerator when using the Quotient Rule?
  4. Can you differentiate sin(t)/t5\sin(t)/t^5 using a similar process?
  5. How would the derivative change if the function were 1t5\frac{1}{t^5}?

Tip: Always simplify expressions as much as possible after applying differentiation rules to make them easier to interpret.

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Math Problem Analysis

Mathematical Concepts

Calculus
Differentiation
Quotient Rule

Formulas

Quotient Rule: (d/dt)[u(t)/v(t)] = (v(t) * u'(t) - u(t) * v'(t)) / [v(t)]^2

Theorems

Quotient Rule

Suitable Grade Level

Grades 11-12