/*! 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 31 Find \(d y / d x\). $$ y=\sq... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 31

Find \(d y / d x\). $$ y=\sqrt[4]{1+\cos \left(x^{2}+2 x\right)} $$

Short Answer

Expert verified
\( \frac{dy}{dx} = -\frac{1}{2} \sin(x^2 + 2x)(2x + 2)(1 + \cos(x^2 + 2x))^{-3/4} \).

Step by step solution

01

Identify the function and apply the chain rule

The given function is \( y = (1 + \cos(x^2 + 2x))^{1/4} \). To find the derivative, we'll use the chain rule which states that if \( y = u^n \), then \( \frac{dy}{dx} = n \cdot u^{n-1} \cdot \frac{du}{dx} \). Here, \( u = 1 + \cos(x^2 + 2x) \) and \( n = \frac{1}{4} \).
02

Differentiate the outer function

Differentiate the outer function \((1 + \cos(x^2 + 2x))^{1/4}\) by applying the rule for derivatives of powers: \( \frac{d}{dx}(u^n) = n \cdot u^{n-1} \). Thus, \( \frac{dy}{du} = \frac{1}{4} (1 + \cos(x^2 + 2x))^{-3/4} \).
03

Differentiate the inner function using the chain rule

The inner function is \( u = 1 + \cos(x^2 + 2x) \). The derivative of \( \cos(v) \) is \(-\sin(v) \cdot \frac{dv}{dx} \). Here, \( v = x^2 + 2x \), so we must find \( \frac{dv}{dx} \).
04

Differentiate \( v = x^2 + 2x \)

Differentiate \( x^2 + 2x \) which gives \( \frac{dv}{dx} = 2x + 2 \).
05

Combine derivatives to find \( \frac{du}{dx} \)

Using the derivative of \( \cos(v) \), we have \( \frac{du}{dx} = -\sin(v) \cdot \frac{dv}{dx} = -\sin(x^2 + 2x) \cdot (2x + 2) \).
06

Combine results to find \( \frac{dy}{dx} \)

Substitute \( \frac{dy}{du} \) and \( \frac{du}{dx} \) into the chain rule formula \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \). So, \( \frac{dy}{dx} = \frac{1}{4} (1 + \cos(x^2 + 2x))^{-3/4} \cdot (-\sin(x^2 + 2x) \cdot (2x + 2)) \).
07

Simplify the expression

Simplify \( \frac{dy}{dx} = -\frac{1}{2} \sin(x^2 + 2x) \cdot (2x + 2) \cdot (1 + \cos(x^2 + 2x))^{-3/4} \).

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 91影视!

Key Concepts

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

Understanding the Chain Rule in Calculus
The chain rule is a fundamental concept in calculus, used when differentiating composite functions. A composite function is essentially a function within another function, like peeling layers of an onion. Imagine you have a function inside another function, denoted as \( y = f(g(x)) \). The chain rule allows you to find the derivative by following a systematic approach:
  • First, differentiate the outer function, considering the inner function as a temporary variable.
  • Then, differentiate the inner function separately.
  • Finally, multiply these derivatives together.

Using the example \( y = (1 + \cos(x^2 + 2x))^{1/4} \), we apply the chain rule:
  • The outer function \( f(u) = u^{1/4} \), differentiated as \( f'(u) = \frac{1}{4}u^{-3/4} \).
  • The inner function \( u = 1 + \cos(x^2 + 2x) \), differentiated as \( u' = -\sin(x^2 + 2x) \cdot (2x + 2) \).
By applying both derivatives and multiplying them, we efficiently find \( \frac{dy}{dx} \) using the chain rule. Understanding this process is crucial for solving many calculus problems.
Basics of Derivatives and Their Application
A derivative represents the rate of change of a function concerning its variable. Imagine you are graphing a curve, and you want to know how steep it is at a certain point; that is precisely what a derivative tells you.
  • It helps in understanding rates of change in real-world phenomena, like speed or growth.
  • In calculus, derivatives let us find slopes, optimize functions, and model natural processes.

In our example function \( y = (1 + \cos(x^2 + 2x))^{1/4} \):
  • The outer derivative \( \frac{dy}{du} = \frac{1}{4}(1 + \cos(x^2 + 2x))^{-3/4} \), shows how \( y \) changes as its outer function changes.
  • Apply derivatives to each stage, beginning with the simplest internal part and gradually to the entire function.
  • The inner derivative \( \frac{du}{dx} = -\sin(x^2 + 2x) \cdot (2x + 2) \), reveals the rate of change concerning \( x \) itself.
These derivative calculations combine to give a full picture of how the function behaves.
Exploring Trigonometric Functions in Calculus
Trigonometric functions like sine, cosine, and tangent have special properties used widely in calculus for multiple applications. These functions model cyclical situations like waves and oscillations.
  • Cosine and sine functions are derivatives of each other. Cosine's derivative is \(-\sin(x)\).
  • They have derivatives that are often involved in transformations and signal processing.

In our original function, trigonometric properties play a crucial role:
  • The cosine function, \( \cos(x^2 + 2x) \), is part of the inner function \( u \), needing differentiation.
  • Its derivative, \( -\sin(x^2 + 2x) \), combines with the inner function's derivative \( (2x + 2) \) to form the composite chain rule application.
Understanding and working through the trigonometric functions in calculus problems like these lends critical insight into their interaction and application.

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

Let \(f(x)=x \sin x\) (a) Draw the graphs of \(f(x)\) and \(f^{\prime}(x)\) on \([\pi, 6 \pi]\). (b) How many solutions does \(f(x)=0\) have on \([\pi, 6 \pi]\) ? How many solutions does \(f^{\prime}(x)=0\) have on this interval? (c) What is wrong with the following conjecture? If \(f\) and \(f^{\prime}\) are both continuous and differentiable on \([a, b],\) if \(f(a)=f(b)=0,\) and if \(f(x)=0\) has exactly \(n\) solutions on \([a, b],\) then \(f^{\prime}(x)=0\) has exactly \(n-1\) solutions on \([a, b]\) (d) Determine the maximum value of \(\left|f(x)-f^{\prime}(x)\right|\) on \([\pi, 6 \pi]\)

Use a CAS to do problems. Draw the graphs of \(f(x)=\cos x-\sin (x / 2)\) and its derivative \(f^{\prime}(x)\) on the interval [0,9] using the same axes. (a) Where on this interval is \(f^{\prime}(x)>0 ?\) (b) Where on this interval is \(f(x)\) increasing as \(x\) increases? (c) Make a conjecture. Experiment with other intervals and other functions to support this conjecture.

Find \(D_{x} y .\) \(y=\csc x=1 / \sin x\)

First find and simplify $$\frac{\Delta y}{\Delta x}=\frac{f(x+\Delta x)-f(x)}{\Delta x}$$ Then find \(d y / d x\) by taking the limit of your answer as \(\Delta x \rightarrow 0 .\) $$ y=x^{2} $$

Consider \(y=x^{3}-1\). (a) Sketch its graph as carefully as you can. (b) Draw the tangent line at (2,7) . (c) Estimate the slope of this tangent line. (d) Calculate the slope of the secant line through (2,7) and \(\left(2.01,(2.01)^{3}-1.0\right)\) (e) Find by the limit process (see Example 1) the slope of the tangent line at (2,7)
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Geometry

Read Explanation

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Statistics

Read Explanation
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.