/*! 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 12 In Exercises \(1-18,\) find \(d ... [FREE SOLUTION] | 91Ó°ÊÓ

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In Exercises \(1-18,\) find \(d y / d x\) $$ y=\frac{\cos x}{1+\sin x} $$

Short Answer

Expert verified
\( \frac{dy}{dx} = \frac{-\sin x - 1}{(1+\sin x)^2} \)

Step by step solution

01

Recognize the Quotient Rule

The function provided is a quotient of two functions: the numerator is \( \cos x \) and the denominator is \( 1 + \sin x \). We will use the quotient rule for differentiation which states \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \).
02

Identify Functions and Their Derivatives

Identify \( u = \cos x \) and \( v = 1 + \sin x \). The derivative of \( u \) with respect to \( x \) is \( \frac{du}{dx} = -\sin x \). The derivative of \( v \) with respect to \( x \) is \( \frac{dv}{dx} = \cos x \).
03

Apply the Quotient Rule

Substitute \( u \), \( v \), and their derivatives into the quotient rule: \[ \frac{dy}{dx} = \frac{(1+\sin x)(-\sin x) - (\cos x)(\cos x)}{(1+\sin x)^2} \].
04

Simplify the Numerator

Simplify the expression in the numerator: \[ (1+\sin x)(-\sin x) = -\sin x - \sin^2 x \] and \( (\cos x)(\cos x) = \cos^2 x \). Combine these to get the numerator as \(-\sin x - \sin^2 x - \cos^2 x\).
05

Further Simplify Using Trigonometric Identity

Use the identity \( \sin^2 x + \cos^2 x = 1 \) to simplify the expression: \[ -\sin x - (\sin^2 x + \cos^2 x) = -\sin x - 1 \].
06

Write the Final Derivative

Place the simplified numerator over the denominator squared: \[ \frac{dy}{dx} = \frac{-\sin x - 1}{(1+\sin x)^2} \].

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Key Concepts

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

Differentiation
Differentiation is the process used to find the rate at which a function changes. It allows us to compute the derivative, which is essentially the slope of the function at a given point. For students learning calculus, getting comfortable with differentiation is essential.
One common method in differentiation, particularly when dealing with ratios of functions, is the quotient rule. This rule is ideal when we have one function divided by another. For instance, in the exercise, we used the quotient rule to differentiate the function \( y = \frac{\cos x}{1+\sin x} \).
  • Recognize the parts of the quotient: numerator and denominator.
  • The quotient rule formula is \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2} \).
  • Compute the derivatives of the numerator and the denominator separately.
In this function, understanding each part of the expression's behavior separately allows us to apply the proper operations to find the derivative effectively.
Trigonometric Functions
Trigonometric functions like sine and cosine are pivotal in calculus and represent periodic phenomena very well. Knowing their properties and derivatives is crucial for solving calculus problems that involve these functions.
In the given exercise, \( \cos x \) and \( 1 + \sin x \) are our primary trigonometric components. Each brings its own derivatives:
  • The derivative of \( \cos x \) is \( -\sin x \).
  • The derivative of \( \sin x \) is \( \cos x \).
Trigonometric identities also play a huge role in simplifying expressions. A common identity used is \( \sin^2 x + \cos^2 x = 1 \). In our exercise, identifying this identity helped simplify the expression of the derivative.
Catching these identities will make problems more approachable and straightforward to solve.
Simplification
After differentiating, simplification helps in obtaining a clear and concise answer. It often involves both algebraic manipulation and trigonometric identities.
In this exercise, after applying the quotient rule, we simplified the expression for a cleaner result. Here's how the simplification unfolded:
  • First, simplify the numerator without changing its value.
  • Use trigonometric identities like \( \sin^2 x + \cos^2 x = 1 \), to replace existing terms.
  • Combine like terms, to reduce complexity.
By replacing \( \sin^2 x + \cos^2 x \) with 1, we simplified the expression in the derivative to get \( -\sin x - 1 \). Putting the simplified numerator over the squared denominator resulted in the final derivative: \( \frac{-\sin x - 1}{(1+\sin x)^2} \).
Mastering simplification is key in calculus to reveal the core results amidst complex processes.

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Most popular questions from this chapter

Cardiac output In the late 1860 \(\mathrm{s}\) , Adolf Fick, a professor of physiology in the Faculty of Medicine in Wurzberg, Germany, developed one of the methods we use today for measuring how much blood your heart pumps in a minute. Your cardiac output as you read this sentence is probably about 7 \(\mathrm{L} / \mathrm{min.}\) At rest it is likely to be a bit under 6 \(\mathrm{L} / \mathrm{min.}\) If you are a trained marathon runner running a marathon, your cardiac output can be as high as 30 \(\mathrm{L} / \mathrm{min} .\) $$\begin{array}{c}{\text { Your cardiac output can be calculated with the formula }} \\ {y=\frac{Q}{D}}\end{array}$$ where \(Q\) is the number of milliliters of \(\mathrm{CO}_{2}\) you exhale in a minute and \(D\) is the difference between the \(\mathrm{CO}_{2}\) concentration \((\mathrm{ml} / \mathrm{L})\) in the blood pumped to the lungs and the \(\mathrm{CO}_{2}\) concentration in the blood returning from the lungs. With \(Q=233 \mathrm{ml} / \mathrm{min}\) and \(D=97-56=41 \mathrm{ml} / \mathrm{L},\) $$y=\frac{233 \mathrm{ml} / \min }{41 \mathrm{ml} / \mathrm{L}} \approx 5.68 \mathrm{L} / \mathrm{min},$$ fairly close to the 6 \(\mathrm{L} / \mathrm{min}\) that most people have at basal (resting conditions. (Data courtesy of J. Kenneth Herd, M.D. Quillan College of Medicine, East Tennessee State University.) Suppose that when \(Q=233\) and \(D=41,\) we also know that \(D\) is decreasing at the rate of 2 units a minute but that \(Q\) remains unchanged. What is happening to the cardiac output?

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