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Derive the formula for the derivative with respect to \(x\) of $$ \text { a. } \sec x . \quad \text { b. } \csc x . \quad \text { c. } \cot x $$

Short Answer

Expert verified
\((\sec x)' = \sec x \tan x, (\csc x)' = -\csc x \cot x, (\cot x)' = -\csc^2 x\).

Step by step solution

01

Recall Basic Identities

To find the derivatives of \( \sec x \), \( \csc x \), and \( \cot x \), begin by recalling their definitions in terms of sine and cosine: \( \sec x = \frac{1}{\cos x} \), \( \csc x = \frac{1}{\sin x} \), and \( \cot x = \frac{\cos x}{\sin x} \).
02

Derivative of Secant (\(\sec x\))

Use the quotient rule to differentiate \( \sec x = \frac{1}{\cos x} \). Let \( u = 1 \) and \( v = \cos x \). The quotient rule states \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \).Calculate the derivatives: \( u' = 0 \) and \( v' = -\sin x \).\[ (\sec x)' = \frac{0 \cdot \cos x - 1 \cdot (-\sin x)}{\cos^2 x} = \frac{\sin x}{\cos^2 x} = \sec x \tan x \]
03

Derivative of Cosecant (\(\csc x\))

Use the quotient rule to differentiate \( \csc x = \frac{1}{\sin x} \). Let \( u = 1 \) and \( v = \sin x \). The quotient rule states \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \).Calculate the derivatives: \( u' = 0 \) and \( v' = \cos x \).\[ (\csc x)' = \frac{0 \cdot \sin x - 1 \cdot \cos x}{\sin^2 x} = -\frac{\cos x}{\sin^2 x} = -\csc x \cot x \]
04

Derivative of Cotangent (\(\cot x\))

Use the quotient rule to differentiate \( \cot x = \frac{\cos x}{\sin x} \). Let \( u = \cos x \) and \( v = \sin x \). The quotient rule states \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \).Calculate the derivatives: \( u' = -\sin x \) and \( v' = \cos x \).\[ (\cot x)' = \frac{(-\sin x) \cdot \sin x - \cos x \cdot \cos x}{\sin^2 x} = \frac{-\sin^2 x - \cos^2 x}{\sin^2 x} = \frac{-1}{\sin^2 x} = -\csc^2 x \]
05

Conclusion

The derivatives are:For \( \sec x \), the derivative is \( \sec x \tan x \).For \( \csc x \), the derivative is \(-\csc x \cot x \).For \( \cot x \), the derivative is \(-\csc^2 x \).

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Key Concepts

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

calculus
Calculus is a branch of mathematics focused on the study of change, just like geometry focuses on shapes and algebra on equations. In calculus, we investigate how things change and grow. The core components of calculus are differentiation and integration.

Differentiation helps us find the rate at which things change. For example, it tells us how fast your height is increasing over time. On the other hand, integration is about accumulation. How much water has filled a container up to a certain point?

Trigonometric derivatives are a particular focus within calculus. They play a crucial role, especially when dealing with angles and periodic phenomena such as waves. Understanding how the derivatives of trigonometric functions like sine, cosine, or tangent behave is important in fields like physics and engineering.
differentiation
Differentiation in calculus refers to the process of finding a derivative. The derivative of a function tells us how the function's output changes when we slightly adjust its input. It's like finding the slope of a curve at a specific point.

When differentiating trigonometric functions such as secant (\( \sec x \)), cosecant (\( \csc x \)), and cotangent (\( \cot x \)), we can use several rules and identities. These help break down complex functions into simpler pieces.

The derivative of \( \sec x \) involves using the quotient rule. Through differentiation, we discover that \( (\sec x)' = \sec x \tan x \). Similarly, for \( \csc x \), using differentiation, we find \( (\csc x)' = -\csc x \cot x \). Understanding these derivatives is crucial for solving various calculus problems.
trigonometric identities
Trigonometric identities are equations involving trigonometric functions that are true for every value of the involved variable. They're like shortcuts that make working with these functions easier.

Some of the most important trigonometric identities include Pythagorean identities, which help us transform and simplify expressions involving sine, cosine, and tangent. For instance, we often use the identities \( \sec x = \frac{1}{\cos x} \), \( \csc x = \frac{1}{\sin x} \), and \( \cot x = \frac{\cos x}{\sin x} \). These identities are essential in tackling the derivatives of these functions.

Using these definitions, we apply the quotient rule from calculus to differentiate functions like \( \frac{1}{\cos x} \) and \( \frac{1}{\sin x} \). This allows us to simplify and solve problems involving trigonometric derivatives with ease.
quotient rule
The quotient rule is a formula used for finding the derivative of a quotient of two functions. It's like dividing the differentiation process into the top and bottom parts of a fraction.

If you have a function \( f(x) = \frac{u(x)}{v(x)} \), the quotient rule states that the derivative \( f'(x) \) is given by:
  • \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \)
This rule is particularly handy when differentiating trigonometric functions given in ratio forms like \( \sec x = \frac{1}{\cos x} \) or \( \cot x = \frac{\cos x}{\sin x} \).

Using the quotient rule, we found that the derivative of \( \sec x \) is \( \sec x \tan x \), \( \csc x \) becomes \( -\csc x \cot x \), and \( \cot x \) results in \( -\csc^2 x \). This method simplifies the differentiation of complex fractional functions and is a powerful tool in calculus.

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Most popular questions from this chapter

Constant acceleration Suppose that the velocity of a falling body is \(v=k \sqrt{s} \mathrm{m} / \mathrm{sec}(k\) a constant) at the instant the body has fallen \(s \mathrm{m}\) from its starting point. Show that the body's acceleration is constant.

Slopes on the graph of the cotangent function Graph \(y=\cot x\) and its derivative together for \(0< x <\pi .\) Does the graph of the cotangent function appear to have a smallest slope? A largest slope? Is the slope ever positive? Give reasons for your answers.

A draining hemispherical reservoir Water is flowing at the rate of 6 \(\mathrm{m}^{3} / \mathrm{min}\) from a reservoir shaped like a hemispherical bowl of radius 13 \(\mathrm{m}\) , shown here in profile. Answer the following questions, given that the volume of water in a hemispherical bowl of radius \(R\) is \(V=(\pi / 3) y^{2}(3 R-y)\) when the water is \(y\) meters deep. a. At what rate is the water level changing when the water is 8 \(\mathrm{m}\) deep? b. What is the radius \(r\) of the water's surface when the water is y \(\mathrm{m}\) deep? c. At what rate is the radius \(r\) changing when the water is 8 \(\mathrm{m}\) deep?

Radians versus degrees: degree mode derivatives What happens to the derivatives of \(\sin x\) and \(\cos x\) if \(x\) is measured in degrees instead of radians? To find out, take the following steps. a. With your graphing calculator or computer grapher in degree mode, graph $$ f(h)=\frac{\sin h}{h} $$ and estimate \(\lim _{h \rightarrow 0} f(h) .\) Compare your estimate with \(\pi / 180 .\) Is there any reason to believe the limit should be \(\pi / 180 ?\) b. With your grapher still in degree mode, estimate $$ \lim _{h \rightarrow 0} \frac{\cos h-1}{h} $$ c. Now go back to the derivation of the formula for the derivative of sin \(x\) in the text and carry out the steps of the derivation using degree-mode limits. What formula do you obtain for the derivative? d. Work through the derivation of the formula for the derivative of cos \(x\) using degree-mode limits. What formula do you obtain for the derivative? e. The disadvantages of the degree-mode formulas become apparent as you start taking derivatives of higher order. Try it. What are the second and third degree-mode derivatives of sin \(x\) and \(\cos x ?\)

Consider the function $$f(x)=\left\\{\begin{array}{cc}{x^{2} \cos \left(\frac{2}{x}\right),} & {x \neq 0} \\ {0,} & {x=0}\end{array}\right.$$ a. Show that \(f\) is continuous at \(x=0\) . b. Determine \(f^{\prime}\) for \(x \neq 0\) . c. Show that \(f\) is differentiable at \(x=0\) . d. Show that \(f^{\prime}\) is not continuous at \(x=0\)

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