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Chapter 3: Problem 61

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

Short Answer

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

Step by step solution

01

Identify the Functions f(x) and g(x) and Calculate their Derivatives

First, identify the functions: $$f(x) = 1 - \cos x$$ $$g(x) = 1 + \cos x$$ Next, find the derivatives of these functions: $$f'(x) = -\sin x$$ $$g'(x) = -\sin x$$
02

Apply the Quotient Rule to Find the Derivative

Using the quotient rule formula, substitute the functions and their derivatives: $$\frac{dy}{dx} = \frac{[-\sin x(1 + \cos x)] - [(1 - \cos x)(-\sin x)]}{(1 + \cos x)^2}$$
03

Simplify the Expression

Simplify the expression in the numerator: $$\frac{dy}{dx} = \frac{-\sin x - \sin x\cos x + \sin x - \sin x\cos x}{(1 + \cos x)^2}$$ Combine the like terms: $$\frac{dy}{dx} = \frac{-2\sin x\cos x}{(1 + \cos x)^2}$$ The derivative of the given function is: $$\frac{dy}{dx} = \frac{-2\sin x\cos x}{(1 + \cos x)^2}$$

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Key Concepts

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

Derivative
A derivative measures how a function changes as its input changes. It's like asking, "How fast is this function moving at any point?" Knowing how to find the derivative of a function is essential in calculus, as it helps us understand rates of change and slopes of curves. When you're given a function like \( y = \frac{1 - \cos x}{1 + \cos x} \), finding its derivative \( \frac{dy}{dx} \) means identifying how the output \( y \) changes as \( x \) changes slightly.
The derivative provides insights such as the steepness of a curve at any point, and it can even tell us if the function is increasing or decreasing.
  • To compute derivatives of complex functions, rules such as the Quotient Rule are used.
  • These rules help in systematically finding derivatives without needing to manually test each function with tiny changes in \( x \).
  • Understanding derivatives is foundational for more advanced calculus topics such as integration and differential equations.
Trigonometric Functions
Trigonometric functions like \( \sin(x) \) and \( \cos(x) \) are essential components in calculus, as they describe relationships involving angles and lengths within triangles. These functions are crucial because they appear frequently in physics, engineering, and calculus.
In our problem, both \( f(x) = 1 - \cos x \) and \( g(x) = 1 + \cos x \) are expressions involving \( \cos(x) \).
When calculating derivatives, the trigonometric identities and their derivatives are frequently used.
  • The derivative of \( \cos(x) \) is \( -\sin(x) \).
  • Understanding how these functions behave over different intervals can help predict the behavior of more complex functions.
  • They also help in simplifying expressions involving trigonometric terms, as seen in the solution by reducing the terms \( -2\sin x\cos x \).
Calculus
Calculus is the mathematical study of change and motion, consisting of two main branches: differential and integral calculus. Differential calculus focuses on derivatives and their properties. By learning calculus, one gains the tools to solve real-world problems involving rates and changes.
When tackling the problem of finding \( dy/dx \) for the given function, you utilize concepts from calculus to break down the problem systematically.
  • The Quotient Rule is a specific technique within differential calculus used for finding the derivative of the division of two functions.
  • Calculus not only helps in solving mathematical problems but also in understanding phenomena in natural sciences, economics, and beyond.
  • It provides a framework for modeling and understanding the continuous change and the accumulation of quantities.
Understanding the basic principles of calculus will empower you to tackle a variety of mathematical and real-world challenges with confidence.

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Most popular questions from this chapter

Consider the following functions (on the given interval, if specified). Find the inverse function, express it as a function of \(x,\) and find the derivative of the inverse function. $$f(x)=\frac{x}{x+5}$$

Consider the following functions (on the given interval, if specified). Find the inverse function, express it as a function of \(x,\) and find the derivative of the inverse function. $$f(x)=x^{2 / 3}, \text { for } x>0$$

One of the Leibniz Rules One of several Leibniz Rules in calculus deals with higher-order derivatives of products. Let \((f g)^{(n)}\) denote the \(n\) th derivative of the product \(f g,\) for \(n \geq 1\) a. Prove that \((f g)^{(2)}=f^{\prime \prime} g+2 f^{\prime} g^{\prime}+f g^{\prime \prime}\) b. Prove that, in general,$$(f g)^{(n)}=\sum_{k=0}^{n}\left(\begin{array}{l} n \\\k\end{array}\right) f^{(k)} g^{(n-k)}$$ where \(\left(\begin{array}{l}n \\\ k\end{array}\right)=\frac{n !}{k !(n-k) !}\) are the binomial coefficients. c. Compare the result of (b) to the expansion of \((a+b)^{n}\).

Vertical tangent lines a. Determine the points at which the curve \(x+y^{3}-y=1\) has a vertical tangent line (see Exercise 52 ). b. Does the curve have any horizontal tangent lines? Explain.

Use the following table to find the given derivatives. $$\begin{array}{llllll} x & 1 & 2 & 3 & 4 & 5 \\ \hline f(x) & 5 & 4 & 3 & 2 & 1 \\ f^{\prime}(x) & 3 & 5 & 2 & 1 & 4 \\ g(x) & 4 & 2 & 5 & 3 & 1 \\ g^{\prime}(x) & 2 & 4 & 3 & 1 & 5 \end{array}$$ $$\left.\frac{d}{d x}(f(x) g(x))\right|_{x=1}$$
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