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\(f(x)=\tan x, \quad a=\pi\)

Short Answer

Expert verified
The derivative of the function \(f(x)=\tan x\) at \(x=\pi\) is 1.

Step by step solution

01

Calculation of Derivative

The derivative of \(f(x)=\tan x\) is \(f'(x)=\sec^2 x\). This formula comes from the standard differentiation rules.
02

Substituting \(x=a=\pi\)

On substituting \(x=\pi\) in \(f'(x)=\sec^2 x\), we get \(f'(\pi)=\sec^2 \pi\).
03

Evaluation of \(\sec^2 \pi\)

Here, \(\sec \pi = -1\) as secant function is reciprocal of cosine function and \(\cos \pi = -1\). Therefore, \(\sec^2 \pi = (-1)^2 = 1\).

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Key Concepts

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

Differentiation Rules
Differentiation is a fundamental concept in calculus that deals with finding the rate at which a function is changing at any given point. It is the process of determining the derivative of a function. There are several rules for differentiation which simplifies this process for a variety of functions. Some of these rules include the power rule, product rule, quotient rule, and chain rule.

For trigonometric functions like the tangent function, specific rules apply. The derivative of a tangent function is not immediately obvious like a simple polynomial, but using the differentiation rules for trigonometric functions, we can determine that the derivative of \(\tan x\) is \(\sec^2 x\). This rule is crucial as it helps to solve a wide array of problems involving the tangent function in calculus.
Secant Function
The secant function, denoted as \(\sec x\), is one of the six basic trigonometric functions. It is defined as the reciprocal of the cosine function, essentially as \(\sec x = \frac{1}{\cos x}\). The secant function has its own unique graph, which illustrates that it possesses both positive and negative values and has undefined points where the cosine function equals zero.

The significance of the secant function in differentiation comes from the fact that it represents the slope of the tangent to the curve of the function \(y=\tan x\) at any point \(x\). Thus, understanding the secant function is critical for grasping the concepts involved in finding derivatives of trigonometric functions.
Reciprocal Trigonometric Functions
In trigonometry, the reciprocal trigonometric functions are those that invert the primary trigonometric functions, namely sine, cosine, and tangent. These reciprocals are the cosecant (\(\csc x = \frac{1}{\sin x}\)), secant (\(\sec x = \frac{1}{\cos x}\)), and cotangent (\(\cot x = \frac{1}{\tan x}\)) functions.

Understanding these reciprocal relationships is particularly useful in calculus when evaluating the derivatives of trigonometric functions. Their properties and the way they behave around important angles (like \(0\), \(\frac{\pi}{2}\), and \(\pi\)) are essential to correctly solving differentiation problems that involve these functions. For instance, the derivative of the tangent function involves the secant squared function, which showcases the importance of reciprocal functions in differentiation.
Evaluating Trigonometric Functions
Evaluating trigonometric functions involves finding the value of these functions at specific angles. Trigonometric functions take an angle as an input and produce a ratio as output. Understanding the values of trig functions at key angles is essential in calculus and trigonometry.

For example, knowing that \(\cos \pi = -1\) helps you quickly find that \(\sec \pi = -1\) since they are reciprocals of each other. This knowledge is applied in evaluating the derivative of the tangent function at \(\pi\), as seen in the solution for the derivative of \(\tan x\). Errors in evaluating these functions at particular angles can lead to mistakes in further calculations, making this skill critical for students to master.

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Most popular questions from this chapter

Motion along a Circle A wheel of radius 2 ft makes 8 revolutions about its center every second. (a) Explain how the parametric equations \(x=2 \cos \theta, \quad y=2 \sin \theta\) \(x=2 \cos \theta, \quad y=2 \sin \theta\) (b) Express \(\theta\) as a function of time \(t\) . (c) Find the rate of horizontal movement and the rate of vertical movement of a point on the edge of the wheel when it is at the position given by \(\theta=\pi / 4, \pi / 2,\) and \(\pi .\)

Motion on a Line The positions of two particles on the \(s\) -axis are \(s_{1}=\sin t\) and \(s_{2}=\sin (t+\pi / 3),\) with \(s_{1}\) and \(s_{2}\) in meters and \(t\) in seconds. (a) At what time \((\mathrm{s})\) in the interval \(0 \leq t \leq 2 \pi\) do the particles meet? (b) What is the farthest apart that the particles ever get? (c) When in the interval \(0 \leq t \leq 2 \pi\) is the distance between the particles changing the fastest?

You may use a graphing calculator to solve the following problems. True or False Newton's method will not find the zero of \(f(x)=x /\left(x^{2}+1\right)\) if the first guess is greater than \(1 .\) Justify your answer.

Multiple Choice Which of the following conditions would enable you to conclude that the graph of \(f\) has a point of inflection at \(x=c ?\) (A) There is a local maximum of \(f^{\prime}\) at \(x=c\) . (B) \(f^{\prime \prime}(c)=0 .\) (C) \(f^{\prime \prime}(c)\) does not exist. (D) The sign of \(f^{\prime}\) changes at \(x=c\) . (E) \(f\) is a cubic polynomial and \(c=0\)

Airplane Landing Path An airplane is flying at altitude \(H\) when it begins its descent to an airport runway that is at horizontal ground distance \(L\) from the airplane, as shown in the figure. Assume that the landing path of the airplane is the graph of a cubic polynomial function \(y=a x^{3}+b x^{2}+c x+d\) where \(y(-L)=H\) and \(y(0)=0 .\) (a) What is \(d y / d x\) at \(x=0 ?\) (b) What is \(d y / d x\) at \(x=-L ?\) (c) Use the values for \(d y / d x\) at \(x=0\) and \(x=-L\) together with \(y(0)=0\) and \(y(-L)=H\) to show that $$y(x)=H\left[2\left(\frac{x}{L}\right)^{3}+3\left(\frac{x}{L}\right)^{2}\right]$$

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