Textbook Question
Derivatives and tangent lines
a. For the following functions and values of a, find f′(a).
f(x) = 8x; a = −3
Ch. 3 - Derivatives
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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Table of contents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 3, Problem 3.5.74a
Find an equation of the line tangent to the following curves at the given value of x.
y = csc x; x = π/4
Verified step by step guidance1
First, understand that the equation of a tangent line to a curve at a given point is y = mx + b, where m is the slope of the tangent line and b is the y-intercept.
To find the slope of the tangent line, calculate the derivative of the function y = csc(x). The derivative of csc(x) is -csc(x)cot(x).
Evaluate the derivative at the given point x = π/4. Substitute x = π/4 into the derivative -csc(x)cot(x) to find the slope m.
Next, find the y-coordinate of the point on the curve by substituting x = π/4 into the original function y = csc(x). This gives you the point (π/4, y).
Use the point-slope form of the equation of a line, y - y₁ = m(x - x₁), where (x₁, y₁) is the point on the curve and m is the slope, to write the equation of the tangent line.
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Video duration:
5mKey Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Tangent Line
A tangent line to a curve at a given point is a straight line that touches the curve at that point without crossing it. The slope of the tangent line is equal to the derivative of the function at that point. This concept is crucial for finding the equation of the tangent line, as it provides the necessary slope to use in the point-slope form of a line.
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Slopes of Tangent Lines
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 that provides the slope of the tangent line at any point on the curve. For the function y = csc x, the derivative can be calculated using differentiation rules, which will be needed to find the slope at x = π/4.
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Cosecant Function
The cosecant function, denoted as csc x, is the reciprocal of the sine function, defined as csc x = 1/sin x. Understanding the properties and behavior of the cosecant function is essential for evaluating the function and its derivative at specific points, such as x = π/4, which is necessary for determining the equation of the tangent line.
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Related Practice
Textbook Question
Derivatives and tangent lines
a. For the following functions and values of a, find f′(a).
f(x) = √2x+1; a= 4
Textbook Question
Derivatives of inverse functions from a table Use the following tables to determine the indicated derivatives or state that the derivative cannot be determined. <IMAGE>
a. (f^-1)'(4)
Textbook Question
Use the definition of the derivative to determine d/dx(ax²+bx+c), where a, b, and c are constants.
Textbook Question
60–62. {Use of Tech} Multiple tangent lines Complete the following steps. <IMAGE>
a. Find equations of all lines tangent to the curve at the given value of x.
x+y³−y=1; x=1
Textbook Question
45–50. Tangent lines Carry out the following steps. <IMAGE>
a. Verify that the given point lies on the curve.
x⁴-x²y+y⁴=1; (−1, 1)