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Chapter 3: Problem 12

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

Short Answer

Expert verified
The derivative is \( \frac{d y}{d x} = \frac{-1 - \sin x}{(1+\sin x)^2} \).

Step by step solution

01

Recognize the Function Type

The given function \( y = \frac{\cos x}{1+\sin x} \) is a rational function, where both the numerator and the denominator are functions of \( x \). This suggests the use of the quotient rule for differentiation.
02

Recall the Quotient Rule Formula

The quotient rule states that if \( y = \frac{u}{v} \), where \( u \) and \( v \) are functions of \( x \), then \( \frac{d y}{d x} = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v^2} \). In this problem, set \( u = \cos x \) and \( v = 1 + \sin x \).
03

Differentiate the Numerator

Differentiate \( u = \cos x \) with respect to \( x \). The derivative of \( \cos x \) is \( -\sin x \). So, \( \frac{d u}{d x} = -\sin x \).
04

Differentiate the Denominator

Differentiate \( v = 1 + \sin x \) with respect to \( x \). The derivative of \( \sin x \) is \( \cos x \), and the derivative of a constant (1) is 0. Thus, \( \frac{d v}{d x} = \cos x \).
05

Apply the Quotient Rule

Substitute the derivatives from Steps 3 and 4 into the quotient rule formula: \[ \frac{d y}{d x} = \frac{(1+\sin x)(-\sin x) - (\cos x)(\cos x)}{(1+\sin x)^2} \] Simplify the expression in the numerator: \[ -(\sin x + \sin^2 x) - \cos^2 x = -\sin x - \sin^2 x - \cos^2 x \] Recall the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \). This reduces to: \[ -\sin x - 1 \] Thus, the derivative is: \[ \frac{d y}{d x} = \frac{-1 - \sin x}{(1+\sin x)^2} \]
06

Simplify the Expression

The derivative \( \frac{d y}{d x} = \frac{-1 - \sin x}{(1+\sin x)^2} \) is already in its simplest form, with the numerator combining all like terms and the identity-based simplification complete.

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Key Concepts

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

Quotient Rule
When solving differentiation problems involving divisions of functions, like the one where you have \( y = \frac{\cos x}{1+\sin x} \), the quotient rule is very handy. The quotient rule helps us find the derivative of the quotient of two functions, that is, when one function is divided by another.
  • Formula: Given two functions \( u(x) = \cos x \) and \( v(x) = 1+\sin x \), the derivative is calculated as follows:\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
  • Here, \( du/dx \) is the derivative of \( u \) with respect to \( x \) and \( dv/dx \) is the derivative of \( v \) with respect to \( x \).
It's crucial to understand each part of this formula 鈥� top and bottom. Substitute your terms in carefully and simplify the results.
By using the quotient rule correctly, you find the rate of change of the function relative to time or another variable.
Trigonometric Functions
Trigonometric functions like \( \cos x \) and \( \sin x \) are frequently encountered in calculus.
  • The sine function \( \sin x \) provides the y-coordinate of a point on the unit circle. Its derivative is \( \cos x \), linking these two functions intricately.
  • The cosine function \( \cos x \) offers the x-coordinate of a point on the unit circle. Its derivative is \(-\sin x\).
When applying the quotient rule to our specific problem, make a mental note of these derivatives.
For the numerator, the derivative of \( \cos x \) translates to \( -\sin x \). For the denominator, \( 1 + \sin x \), the derivative becomes \( \cos x \) because the derivative of a constant is zero. Knowing these derivatives is key to tackling the differentiation task efficiently.
Calculus
Calculus allows us to explore and understand a wide range of phenomena that involve change. Differentiation, a fundamental concept in calculus, helps us determine how a function changes at any given point.
  • It's about finding the slope of the function's graph at any point 鈥� essentially, how steep it is.
  • Tools like the quotient rule and knowledge of trigonometric derivatives expand our ability to handle complex functions.
In the given exercise, you're effectively using calculus to break down the behavior of a function involving trigonometric terms.
By applying calculus, you go beyond just understanding the function's value at a point and delve into the change and motion it represents. This mathematical insight extends to numerous real-world applications like physics, engineering, and economics, where understanding change or motion is crucial.

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Most popular questions from this chapter

Find \(d y / d t\) $$y=\sec ^{2} \pi t$$

A weight is attached to a spring and reaches its equilibrium position \((x=0) .\) It is then set in motion resulting in a displacement of $$x=10 \cos t$$ where \(x\) is measured in centimeters and \(t\) is measured in seconds. See the accompanying figure. a. Find the spring's displacement when \(t=0, t=\pi / 3,\) and \(t=3 \pi / 4\) b. Find the spring's velocity when \(t=0, t=\pi / 3,\) and \(t=3 \pi / 4\)

Use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval \(I\). Perform the following steps: a. Plot the function \(f\) over \(I\) b. Find the linearization \(L\) of the function at the point \(a\) c. Plot \(f\) and \(L\) together on a single graph. d. Plot the absolute error \(|f(x)-L(x)|\) over \(I\) and find its maximum value. e. From your graph in part (d), estimate as large a \(\delta>0\) as you can, satisfying \(|x-a|<\delta \quad \Rightarrow \quad|f(x)-L(x)|<\epsilon\) for \(\epsilon=0.5,0.1,\) and \(0.01 .\) Then check graphically to see if your \(\delta\) -estimate holds true. $$f(x)=\frac{x-1}{4 x^{2}+1}, \quad\left[-\frac{3}{4}, 1\right], \quad a=\frac{1}{2}$$

When the length \(L\) of a clock pendulum is held constant by controlling its temperature, the pendulum's period \(T\) depends on the acceleration of gravity \(g\). The period will therefore vary slightly as the clock is moved from place to place on the earth's surface, depending on the change in \(g.\) By keeping track of \(\Delta T,\) we can estimate the variation in \(g\) from the equation \(T=2 \pi(L / g)^{1 / 2}\) that relates \(T, g,\) and \(L\). a. With \(L\) held constant and \(g\) as the independent variable, calculate \(d T\) and use it to answer parts (b) and (c). b. If \(g\) increases, will \(T\) increase or decrease? Will a pendulum clock speed up or slow down? Explain. c. A clock with a \(100-\mathrm{cm}\) pendulum is moved from a location where \(g=980 \mathrm{cm} / \mathrm{s}^{2}\) to a new location. This increases the period by \(d T=0.001\) s. Find \(d g\) and estimate the value of \(g\) at the new location.

Find the tangent to \(y=((x-1) /(x+1))^{2}\) at \(x=0\)
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