Ch. 3 - Derivatives

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 3 - Derivatives

Problem 3.R.63

Chapter 3, Problem 3.R.63

Evaluate d/dx(x sec^−1 x) |x = 2 /√3.

Verified step by step guidance

First, recognize that you need to differentiate the function f(x) = x sec^(-1)(x) with respect to x. This requires the use of the product rule for differentiation.

Recall the product rule: if you have a function h(x) = u(x) * v(x), then the derivative h'(x) = u'(x) * v(x) + u(x) * v'(x). Here, let u(x) = x and v(x) = sec^(-1)(x).

Differentiate u(x) = x to get u'(x) = 1.

Differentiate v(x) = sec^(-1)(x). The derivative of sec^(-1)(x) with respect to x is 1 / (|x| * sqrt(x^2 - 1)).

Apply the product rule: f'(x) = 1 * sec^(-1)(x) + x * (1 / (|x| * sqrt(x^2 - 1))). Finally, substitute x = 2/√3 into this expression to evaluate the derivative at that point.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Derivative

The derivative of a function measures how the function's output value changes as its input value changes. It is a fundamental concept in calculus, representing the slope of the tangent line to the curve at a given point. The notation d/dx indicates differentiation with respect to x, and finding the derivative is essential for understanding rates of change and optimizing functions.

Secant and Inverse Secant Functions

The secant function, denoted as sec(x), is the reciprocal of the cosine function, defined as sec(x) = 1/cos(x). The inverse secant function, sec^−1(x), returns the angle whose secant is x. Understanding these functions is crucial for evaluating expressions involving them, especially when differentiating products that include inverse trigonometric functions.

Product Rule

The product rule is a formula used to find the derivative of the product of two functions. It states that if u(x) and v(x) are two differentiable functions, then the derivative of their product is given by d/dx[u(x)v(x)] = u'(x)v(x) + u(x)v'(x). This rule is essential for solving the given problem, as it allows for the differentiation of the product x and sec^−1(x).

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The Product Rule

Related Practice

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Textbook Question

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A jet flying at 450 mi/hr and traveling in a straight line at a constant elevation of 500 ft passes directly over a spectator at an air show. How quickly is the angle of elevation (between the ground and the line from the spectator to the jet) changing 2 seconds later?

Textbook Question

Evaluate and simplify y'.

y = ln w / w⁵

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Consider the following functions. In each case, without finding the inverse, evaluate the derivative of the inverse at the given point.

y =√x³+x−1 at y=3

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d/dx (tan³ x-3 tan x+3x) = 3 tan⁴x