/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 Finding a Derivative of a Trigon... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Finding a Derivative of a Trigonometric Function In Exercises \(35-54\) , find the derivative of the trigonometric function. $$y=\cos 4 x$$

Short Answer

Expert verified
The derivative of \(y=\cos(4x)\) is \(-4\sin(4x)\).

Step by step solution

01

Identify the Inner and Outer Functions

In the function \(y=\cos(4x)\), the outer function is the cosine function, and the inner function is \(4x\).
02

Apply the Chain Rule

Based on the chain rule, the derivative of the function is the derivative of the outer function times the derivative of the inner function.
03

Find the Derivative of the Outer Function

The derivative of the cosine function is -sin, so the derivative of \(\cos(4x)\) is \(-\sin(4x)\).
04

Find the Derivative of the Inner Function

The derivative of \(4x\) is 4.
05

Multiply the Derivatives Together

Multiplying the results from Step 3 and Step 4, we get that the derivative of \(y=\cos(4x)\) is \(-4\sin(4x)\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Chain Rule
The chain rule is a fundamental concept in calculus used for finding the derivative of composite functions. When you have a function nested within another, like in our case with \(y = \cos(4x)\), the chain rule helps us take derivatives efficiently.

Here's how it works in simple terms:
  • Identify the outer function, which in many cases is the operation performed last when evaluating the function. For \(y = \cos(4x)\), the outer function is \(\cos(u)\), where \(u = 4x\).
  • Identify the inner function, which is the one you substitute into the outer function. Here, the inner function is \(4x\).
  • The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, times the derivative of the inner function.
So, using the chain rule, the derivative of \(y = \cos(4x)\) becomes \(-\sin(4x) \times 4 = -4\sin(4x)\). This simple multiplication makes the chain rule a powerful tool when dealing with composite functions.
Trigonometric Functions
Trigonometric functions are ratios of sides in right triangles and are fundamental in calculus. Functions like sine, cosine, and tangent are especially important.

In the problem, we’re working with the cosine function, which is frequently encountered in calculus. Here's what you need to understand about cosine:
  • The cosine of an angle in a right triangle is the ratio of the adjacent side to the hypotenuse.
  • Cosine is an even function, meaning \(\cos(-x) = \cos(x)\).
  • It's periodic, repeating every \(2\pi\), which can be particularly useful when dealing with wave patterns.
The derivative properties of trigonometric functions are crucial for applications. For example, the derivative of \(\cos(x)\) is \(-\sin(x)\).

This directly impacts our derivative calculation for \(y = \cos(4x)\), where the formula becomes \(-\sin(4x)\) after applying the chain rule.
Derivatives
Derivatives represent the rate of change or the slope of a function at a certain point. They are one of the cornerstones of calculus, capturing how a function changes as its input changes.

Here's what you need to know about derivatives and their application in our exercise:
  • Derivatives can help understand real-world phenomena, like speed and acceleration, by describing how quickly something is changing.
  • The derivative of a linear function, like \(4x\), is constant because the slope doesn't change. This is why \( \frac{d}{dx}(4x) = 4 \).
  • When we compute the derivative of \(y = \cos(4x)\), each component's derivative from the chain rule provides a comprehensive description of the function’s behavior.
For \(y = \cos(4x)\), the derivative \(-4\sin(4x)\) tells us how \(y\) changes as \(x\) changes, which is the crux of using derivatives.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Finding a Pattern Consider the function \(f(x)=\sin \beta x\) where \(\beta\) is a constant. (a) Find the first-, second-, third-, and fourth-order derivatives of the function. (b) Verify that the function and its second derivative satisfy the equation \(f^{\prime \prime}(x)+\beta^{2} f(x)=0\) (c) Use the results of part (a) to write general rules for the even- and odd- order derivatives \(f^{(2 k)}(x)\) and \(f^{(2 k-1)}(x)\) [Hint: \((-1)^{k}\) is positive if \(k\) is even and negative if \(k\) is odd. \(]\)

Electricity The combined electrical resistance \(R\) of two resistors \(R_{1}\) and \(R_{2},\) connected in parallel, is given by \(\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}\) where \(R, R_{1},\) and \(R_{2}\) are measured in ohms. \(R_{1}\) and \(R_{2}\) are increasing at rates of 1 and 1.5 ohms per second, respectively. At what rate is \(R\) changing when \(R_{1}=50\) ohms and \(R_{2}=75\) ohms?

Wave Motion A buoy oscillates in simple harmonic motion \(y=A \cos \omega t\) as waves move past it. The buoy moves a total of 3.5 feet (vertically) from its low point to its high point. It returns to its high point every 10 seconds. (a) Write an equation describing the motion of the buoy if it is at its high point at \(t=0 .\) (b) Determine the velocity of the buoy as a function of \(t .\)

Think About It Describe the relationship between the rate of change of \(y\) and the rate of change of \(x\) in each expression. Assume all variables and derivatives are positive.

Evaluating a Second Derivative In Exercises \(89-92\) , evaluate the second derivative of the function at the given point. Use a computer algebra system to verify your result. $$f(x)=\frac{1}{\sqrt{x+4}},\left(0, \frac{1}{2}\right)$$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.