/*! 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 23 Find \(d y / d x\) for the follo... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 23

Find \(d y / d x\) for the following functions. $$y=\frac{\cos x}{\sin x+1}$$

Short Answer

Expert verified
Question: Find the derivative of the function \(y=\frac{\cos x}{\sin x+1}\) with respect to x. Answer: The derivative of the given function is \(\frac{dy}{dx}=\frac{-1 - \sin x} {(\sin x +1)^2}\).

Step by step solution

01

Identify the numerator and denominator function

In this exercise, we have \(y=\frac{\cos x}{\sin x+1}\). So, \(u=\cos x\) and \(v=\sin x+1\). We need to differentiate u and v with respect to x.
02

Differentiate the numerator (u) and denominator (v)

Differentiate u with respect to x: $$ \frac{d(\cos x)}{dx} = -\sin x $$ Differentiate v with respect to x: $$ \frac{d(\sin x+1)}{dx} = \cos x $$
03

Apply the quotient rule for differentiation

Now we will use the quotient rule to find the derivative \(y'=\frac{u'v-uv'}{v^2}\). We have \(u' = -\sin x\), \(v' = \cos x\), \(u = \cos x\) and \(v = \sin x + 1\). Substitute the values into the quotient rule: $$ \frac{dy}{dx}=\frac{-\sin x (\sin x +1)-\cos x \cos x} {(\sin x +1)^2} $$
04

Simplify the derivative expression

Expand the terms in the numerator and simplify: $$ \frac{dy}{dx}=\frac{-\sin^2 x - \sin x - \cos ^2 x} {(\sin x +1)^2} $$ Now, we know that \(\sin^2 x + \cos^2 x = 1\), so \(-\sin^2 x - \cos^2 x = -1\). Substitute this into the expression: $$ \frac{dy}{dx}=\frac{-1 - \sin x} {(\sin x +1)^2} $$ Result: The derivative of the given function is $$ \frac{dy}{dx}=\frac{-1 - \sin x} {(\sin x +1)^2} $$

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!

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

Use any method to evaluate the derivative of the following functions. $$f(z)=z^{2}\left(e^{3 z}+4\right)-\frac{2 z}{z^{2}+1}$$

Special Quotient Rule In general, the derivative of a quotient is not the quotient of the derivatives. Find nonconstant functions \(f\) and \(g\) such that the derivative of \(f / g\) equals \(f^{\prime} / g^{\prime}\).

The bottom of a large theater screen is \(3 \mathrm{ft}\) above your eye level and the top of the screen is \(10 \mathrm{ft}\) above your eye level. Assume you walk away from the screen (perpendicular to the screen) at a rate of \(3 \mathrm{ft} / \mathrm{s}\) while looking at the screen. What is the rate of change of the viewing angle \(\theta\) when you are \(30 \mathrm{ft}\) from the wall on which the screen hangs, assuming the floor is horizontal (see figure)?

Find \(f^{\prime}(x), f^{\prime \prime}(x),\) and \(f^{\prime \prime \prime}(x)\) \(f(x)=\frac{1}{x}\)

Gravitational force The magnitude of the gravitational force between two objects of mass \(M\) and \(m\) is given by \(F(x)=-\frac{G M m}{x^{2}},\) where \(x\) is the distance between the centers of mass of the objects and \(G=6.7 \times 10^{-11} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}\) is the gravitational constant (N stands for newton, the unit of force; the negative sign indicates an attractive force). a. Find the instantaneous rate of change of the force with respect to the distance between the objects. b. For two identical objects of mass \(M=m=0.1 \mathrm{kg},\) what is the instantaneous rate of change of the force at a separation of \(x=0.01 \mathrm{m} ?\) c. Does the instantaneous rate of change of the force increase or decrease with the separation? Explain.
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Discrete Mathematics

Read Explanation

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Theoretical and Mathematical Physics

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.