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Chapter 3: Problem 59

Find \(d y / d x\) for the following functions .$$y=\frac{2 \cos x}{1+\sin x}$$

Short Answer

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

Step by step solution

01

Identify the Functions

Identify the functions u(x) and v(x), where .$$u(x)=2\cos x$$ and .$$v(x)=1+\sin x$$
02

Find Derivatives of u(x) and v(x)

Find the derivatives of these functions with respect to x: Using basic differentiation rules, we know that .$$ \frac{d}{d x}\left(\cos x\right) = -\sin x$$ so .$$ \frac{d}{d x}\left(2\cos x\right) = 2(-\sin x) = -2\sin x$$ Thus, .$$ u'(x)= -2\sin x$$ Similarly, .$$ \frac{d}{d x}\left(\sin x\right) = \cos x$$ so .$$ \frac{d}{d x}\left(1+\sin x\right) = 0+\cos x = \cos x$$ Thus, .$$ v'(x)= \cos x$$
03

Apply Quotient Rule

Now apply the quotient rule, which states that if .$$ y = \frac{u(x)}{v(x)}$$ then .$$ \frac{d y}{d x} = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$$ Plug in the values we've found so far for u(x), v(x), u'(x), and v'(x), and we get .$$ \frac{d y}{d x} = \frac{(-2\sin x)(1+\sin x) - (2\cos x)(\cos x)}{(1+\sin x)^2}$$
04

Simplify the Expression

Now, we'll simplify the expression above: $$\frac{d y}{d x} = \frac{-2\sin x - 2\sin^2 x - 2\cos^2 x}{(1+\sin x)^2}$$ Since, .$$ \sin^2 x + \cos^2 x = 1$$ then, .$$ -2\cos^2 x = -2(1 - \sin^2 x) = -2 + 2\sin^2 x$$ Substituting this back into our expression, we get: .$$ \frac{d y}{d x} = \frac{-2\sin x - 2\sin^2 x - (-2 + 2\sin^2 x)}{(1+\sin x)^2}$$ Simplifying further: .$$ \frac{d y}{d x} = \frac{-2\sin x}{(1+\sin x)^2}$$ And that's our final answer for the derivative.

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Key Concepts

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

Differentiation
Differentiation is a fundamental concept in calculus. It involves finding the derivative of a function, which tells us how the function's output changes as its input changes.
To differentiate, we apply rules such as the power rule, product rule, and quotient rule.
In this exercise, we use the **quotient rule** to differentiate a function that is a ratio of two other functions:
  • The **quotient rule** is applied when differentiating functions in the form: \( y = \frac{u(x)}{v(x)} \)
  • The rule states: \( \frac{dy}{dx} = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \)
By systematically identifying and differentiating the numerator \(u(x)\) and denominator \(v(x)\), we apply this formula to find the derivative of the given function.
Trigonometric Functions
Trigonometric functions, such as sine and cosine, often appear in calculus problems.
They have specific rules for differentiation that are essential to know:
  • The derivative of \( \cos x \) is \( -\sin x \).
  • The derivative of \( \sin x \) is \( \cos x \).
In this example, we identified \( u(x) = 2 \cos x \) and \( v(x) = 1 + \sin x \), and then differentiated them:
  • Derivative of \( u(x) = 2 \cos x \) is \( u'(x) = -2 \sin x \).
  • Derivative of \( v(x) = 1 + \sin x \) is \( v'(x) = \cos x \).
These derivatives are crucial for applying the quotient rule correctly.
Understanding each trigonometric function and its derivative allows us to tackle problems involving trigonometric functions effectively.
Simplifying Expressions
After applying the differentiation rules, it鈥檚 important to simplify the resulting expression.
This makes the derivative more readable and helps in understanding the behavior of the function:
  • When simplifying, look out for common trigonometric identities like \( \sin^2 x + \cos^2 x = 1 \).
  • In our solution, the identity \( \cos^2 x = 1 - \sin^2 x \) was used to simplify the expression.
Initially, the derivative expression appeared complex, but through simplification, we arrived at:
\( \frac{d y}{d x} = \frac{-2\sin x}{(1+\sin x)^2} \).
Clear and concise expressions make it easier to analyze and apply the derivative in various contexts.

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Most popular questions from this chapter

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.

Calculate the following derivatives using the Product Rule. $$\begin{array}{lll} \text { a. } \frac{d}{d x}\left(\sin ^{2} x\right) & \text { b. } \frac{d}{d x}\left(\sin ^{3} x\right) & \text { c. } \frac{d}{d x}\left(\sin ^{4} x\right) \end{array}$$ d. Based upon your answers to parts (a)-(c), make a conjecture about \(\frac{d}{d x}\left(\sin ^{n} x\right),\) where \(n\) is a positive integer. Then prove the result by induction.

Derivatives and inverse functions $$\text { Find }\left(f^{-1}\right)^{\prime}(3) \text { if } f(x)=x^{3}+x+1$$

Identifying functions from an equation The following equations implicitly define one or more functions. a. Find \(\frac{d y}{d x}\) using implicit differentiation. b. Solve the given equation for \(y\) to identify the implicitly defined functions \(y=f_{1}(x), y=f_{2}(x), \ldots\) c. Use the functions found in part (b) to graph the given equation. $$y^{2}=\frac{x^{2}(4-x)}{4+x} \text { (right strophoid) }$$

A port and a radar station are 2 mi apart on a straight shore running east and west. A ship leaves the port at noon traveling northeast at a rate of \(15 \mathrm{mi} / \mathrm{hr}\). If the ship maintains its speed and course, what is the rate of change of the tracking angle \(\theta\) between the shore and the line between the radar station and the ship at 12: 30 p.m.? (Hint: Use the Law of sines.)
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