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In Problems 33-40, apply the Chain Rule more than once to find the indicated derivative. $$ D_{t}\left[\cos ^{5}(4 t-19)\right] $$

Short Answer

Expert verified
The derivative is \(-20\cos^4(4t-19)\sin(4t-19).\)

Step by step solution

01

Identify the Composition

The function given is a composition of functions. The outer function is the fifth power, \( u^5 \,,(\text{where }\,u=\cos (4t-19)) \), and the inner function is the cosine of \( 4t-19 \). This helps us understand how the Chain Rule will be applied multiple times.
02

Apply the Chain Rule - First Application

First, apply the Chain Rule to the outer function \( u^5 \). The derivative with respect to \( u \) is \ 5u^4 \, and then multiply this by the derivative of \ u \, which is \( rac{d}{dt}[ ext{cos}(4t-19)] \). So the derivative is: \[ 5 imes ext{cos}^4(4t-19) imes rac{d}{dt}[ ext{cos}(4t-19)]. \]
03

Apply the Chain Rule - Second Application

For the inner function, apply the Chain Rule to \( ext{cos}(4t-19) \). The derivative of \ ext{cos}(x) \ is \ - ext{sin}(x) \, so taking the derivative with respect to \ t \ gives \[ rac{d}{dt}[ ext{cos}(4t-19)] = - ext{sin}(4t-19) imes rac{d}{dt}[4t-19]. \]
04

Differentiate the Linear Function

The derivative of the linear function \( 4t-19 \) with respect to \( t \) is simply the coefficient of \( t \), which is \( 4 \). Hence, \( rac{d}{dt}[4t-19] = 4 \).
05

Combine Derivatives

Substitute the derivative from each operation back into the expression: Substituting the results from the second application of the Chain Rule, we have \[ rac{d}{dt}[ ext{cos}(4t-19)] = - ext{sin}(4t-19) imes 4 = -4 ext{sin}(4t-19).\] Thus, the overall derivative is \[ 5 imes ext{cos}^4(4t-19) imes (-4 ext{sin}(4t-19)). \] Simplifying, we get \[ -20 ext{cos}^4(4t-19) ext{sin}(4t-19). \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Composition of Functions
A fundamental idea in calculus and much of mathematics is the composition of functions. In our exercise, we dealt with a composite function that looked like this: \( \cos^5(4t-19) \). Here, we have layers of functions: an outer function and an inner function.
  • **Outer Function**: The outermost layer is the raising to the fifth power, specifically applied to \( u = \cos(4t-19) \).
  • **Inner Function**: For the inner layer, consider the function \( \cos(x) \) where \( x = 4t-19 \) is another function of \( t \).
Understanding the composition of functions is crucial to applying the Chain Rule effectively, as it helps you see where you need to differentiate at each step. This builds a solid foundation for more complex calculus operations.
Differentiation
Differentiation is the process of finding the derivative of a function, which tells us the rate at which one quantity changes with respect to another. It's a central operation in calculus.
In the given problem, we differentiate a composite function using the Chain Rule. This required multiple steps because we must differentiate both the outer and inner functions. The outer derivative first gives us \( 5\cos^4(4t-19) \). Then we take the derivative of the inner function, which begins with \( \cos(4t-19) \), resulting in an expression that involves \(-\sin(4t-19)\).
Differentiation forms the core of calculus problem solving, as it allows us to find instantaneous rates of change, solve optimization problems, and understand the movement of particles along a path.
Trigonometric Functions
Trigonometric functions like sine and cosine frequently appear in calculus problems, especially when dealing with periodic behaviors or oscillations.
For our exercise, trigonometric functions play a key role in the inner layers of the composite function. The function \( \cos(4t-19) \) signifies the cyclical nature and involves the angle \( 4t-19 \). The derivative of a cosine function is \(-\sin(x)\), showcasing how trigonometric functions can lead to simple yet interesting changes in the rate of angle-related quantities.
Working with trigonometric functions requires familiarity with their properties and transformations, making them vital for a comprehensive understanding of calculus.
Calculus Problem Solving
When solving calculus problems, it is key to have a systematic approach. This involves understanding the problem, breaking it down into manageable parts, and applying appropriate mathematical operations.
In our example, solving the derivative required:
  • Identifying the composition of functions involved.
  • Using the Chain Rule effectively multiple times.
  • Calculating derivatives of trigonometric components.
  • Simplifying the final expression to get the solution \(-20\cos^4(4t-19)\sin(4t-19)\).
Each step is crucial for reaching an accurate solution. Approach calculus problem solving as a journey through layers. Identify outer and inner functions, apply rules thoughtfully, and ensure each differentiation step builds correctly onto the next.

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