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Chapter 3: Problem 39

Find the derivative of the following functions. $$y=\frac{\sin x}{1+\cos x}$$

Short Answer

Expert verified
Question: Find the derivative of the given function: $$y=\frac{\sin x}{1+\cos x}$$ Answer: The derivative of the function is $$y' = \frac{1}{1 + \cos(x)}$$.

Step by step solution

01

Find the derivatives of u(x) and v(x)

Using the properties of derivatives, we can find the derivatives of sine and cosine: u'(x) = derivative of u(x) = derivative of sin(x) = cos(x) v'(x) = derivative of v(x) = derivative of (1 + cos(x)) Since the derivative of 1 (a constant) is 0, and the derivative of cos(x) is -sin(x), we have: v'(x) = 0 - sin(x) = -sin(x)
02

Apply the Quotient Rule

Now, using the Quotient Rule, let's find the derivative of y: $$y' = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$$ Substitute the expressions for u(x), u'(x), v(x), and v'(x) in the formula: $$y' = \frac{\cos(x)(1 + \cos(x)) - \sin(x)(-sin(x))}{(1 + \cos(x))^2}$$
03

Simplify the expression

Let's simplify the expression: $$y' = \frac{\cos(x) + \cos^2(x) + \sin^2(x)}{(1 + \cos(x))^2}$$ We know that $$\sin^2(x) + \cos^2(x) = 1$$, so the numerator becomes: $$1 + \cos(x)$$ Thus, the simplified expression for the derivative is: $$y' = \frac{1 + \cos(x)}{(1 + \cos(x))^2}$$ Now we can further simplify the expression by cancelling a (1 + cos(x)) term from the numerator and the denominator: $$y' = \frac{1}{1 + \cos(x)}$$ So the derivative of the given function $$y=\frac{\sin x}{1+\cos x}$$ is: $$y' = \frac{1}{1 + \cos(x)}$$

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Key Concepts

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

Quotient Rule
The Quotient Rule is a method used in calculus to find the derivative of a function that is the division of two other functions, like \(y = \frac{u(x)}{v(x)}\). It allows us to differentiate complex expressions where one function is divided by another. The formula is:\[y' = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}\]Here's how it works:
  • \(u(x)\) is the function in the numerator
  • \(v(x)\) is the function in the denominator
  • \(u'(x)\) is the derivative of the numerator
  • \(v'(x)\) is the derivative of the denominator
It's essential to remember that you have to square the denominator in the formula. This makes the Quotient Rule particularly helpful when dealing with fractions. In the original problem, \(y = \frac{\sin x}{1+\cos x}\), applying the Quotient Rule led us to simplify and find the derivative easily.
Trigonometric Functions
Trigonometric functions such as \(\sin(x)\) and \(\cos(x)\) are fundamental in calculus. They describe the properties of angles and triangles, and are used frequently in various differentiation problems. Here are some key points:
  • The derivative of \(\sin(x)\) is \(\cos(x)\)
  • The derivative of \(\cos(x)\) is \(-\sin(x)\)
  • Trigonometric identities like \(\sin^2(x) + \cos^2(x) = 1\) can simplify expressions
In the original problem, these derivatives helped us find the necessary changes in \(\sin(x)\) and \(1 + \cos(x)\) while applying the Quotient Rule. Understanding these fundamental derivatives and identities simplifies the differentiation of trigonometric functions considerably.
Differentiation Techniques
Differentiation is the process of finding the derivative and is fundamental to calculus. There are various techniques to perform differentiation effectively, depending on the complexity of the function:
  • **Power Rule**: A basic technique where the derivative of \(x^n\) is \(nx^{n-1}\).
  • **Product Rule**: Used when differentiating products of functions, but in our problem, the focus was on the Quotient Rule.
  • **Quotient Rule**: Essential for differentiating fractions, as seen in the given problem.
  • **Chain Rule**: Useful for composite functions, where one function is inside another.
Each technique has specific applications that simplify the process of finding derivatives. For example, in the problem, the Quotient Rule enabled us to handle the fraction \(y=\frac{\sin x}{1+\cos x}\). By understanding these techniques, you can approach various differentiation challenges confidently.

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Most popular questions from this chapter

Use implicit differentiation to find\(\frac{d y}{d x}.\) $$\sqrt{x+y^{2}}=\sin y$$

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Earth's atmospheric pressure decreases with altitude from a sea level pressure of 1000 millibars (a unit of pressure used by meteorologists). Letting \(z\) be the height above Earth's surface (sea level) in kilometers, the atmospheric pressure is modeled by \(p(z)=1000 e^{-z / 10}\) a. Compute the pressure at the summit of Mt. Everest, which has an elevation of roughly \(10 \mathrm{km} .\) Compare the pressure on Mt. Everest to the pressure at sea level. b. Compute the average change in pressure in the first \(5 \mathrm{km}\) above Earth's surface. c. Compute the rate of change of the pressure at an elevation of \(5 \mathrm{km}\) d. Does \(p^{\prime}(z)\) increase or decrease with \(z ?\) Explain. e. What is the meaning of \(\lim _{z \rightarrow \infty} p(z)=0 ?\)
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