Textbook Question
Find the derivatives of the functions in Exercises 19–40.
q = sin(t / (√t + 1))
Ch. 3 - Derivatives
Hass15th EditionThomas' CalculusISBN: 9780137616077
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Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 3, Problem 3.1.16
In Exercises 11–18, find the slope of the function’s graph at the given point. Then find an equation for the line tangent to the graph there.
h(t) = t³ + 3t, (1, 4)
Verified step by step guidance1
To find the slope of the function's graph at the given point, we need to compute the derivative of the function h(t) = t³ + 3t. The derivative, h'(t), represents the slope of the tangent line at any point t.
Differentiate the function h(t) = t³ + 3t with respect to t. Using the power rule, the derivative of t³ is 3t², and the derivative of 3t is 3. Therefore, h'(t) = 3t² + 3.
Evaluate the derivative at the given point t = 1 to find the slope of the tangent line. Substitute t = 1 into h'(t) to get h'(1) = 3(1)² + 3.
Now that we have the slope of the tangent line, we can use the point-slope form of a line to find the equation of the tangent line. The point-slope form is y - y₁ = m(x - x₁), where m is the slope and (x₁, y₁) is the given point.
Substitute the slope found in step 3 and the point (1, 4) into the point-slope form equation to find the equation of the tangent line. This will give you the equation of the line tangent to the graph at the point (1, 4).
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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 at a point provides the slope of the tangent line to the function's graph at that point. It is calculated as the limit of the average rate of change of the function as the interval approaches zero. For the function h(t) = t³ + 3t, the derivative h'(t) is found using the power rule, resulting in h'(t) = 3t² + 3.
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Derivatives
Tangent Line
A tangent line to a curve at a given point is a straight line that just 'touches' the curve at that point. It has the same slope as the curve at that point, which is given by the derivative. For the function h(t) at the point (1, 4), the slope of the tangent line is h'(1) = 6, and the equation of the tangent line can be found using the point-slope form.
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Point-Slope Form
The point-slope form of a line's equation is useful for writing the equation of a line when you know a point on the line and its slope. It is expressed as y - y₁ = m(x - x₁), where m is the slope and (x₁, y₁) is the point. For the tangent line at (1, 4) with slope 6, the equation is y - 4 = 6(x - 1).
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Related Practice
Textbook Question
Second Derivatives
Find y'' in Exercises 59–64.
y = (1 − √x)⁻¹
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Derivative Calculations
In Exercises 1–8, given y = f(u) and u = g(x), find dy/dx = f'(g(x)) g'(x).
y = sin u, u = x − cos x
Textbook Question
Differentiating Implicitly
Use implicit differentiation to find dy/dx in Exercises 1–14.
x = sec y
Textbook Question
Derivative Calculations
In Exercises 1–8, given y = f(u) and u = g(x), find dy/dx = f'(g(x)) g'(x).
y = sin u, u = 3x + 1
Textbook Question
Find the points on the curve y = tan x, -π/2 < x < π/2, where the normal line is parallel to the line y = -x/2. Sketch the curve and normal lines together, labeling each with its equation.