Textbook Question
Find the derivatives of the functions in Exercises 19–40.
y = (4x + 3)⁴(x + 1)⁻³
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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 3.1.28
Find an equation of the straight line having slope 1/4 that is tangent to the curve y = √x.
Verified step by step guidance1
First, understand that the problem requires finding a tangent line to the curve y = √x with a given slope of 1/4.
To find the point of tangency, differentiate the curve y = √x to find its derivative, which represents the slope of the tangent line at any point x. The derivative is given by: .
Set the derivative equal to the given slope of 1/4 to find the x-coordinate of the point of tangency: = .
Solve the equation from the previous step to find the x-coordinate of the point of tangency. Once you have the x-coordinate, substitute it back into the original curve equation y = √x to find the corresponding y-coordinate.
With the point of tangency (x, y) and the slope 1/4, use the point-slope form of a line to write the equation of the tangent line: y - y₁ = m(x - x₁), where m is the slope and (x₁, y₁) is the point of tangency.
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Key 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 represents the instantaneous rate of change of the function at that point. To find the equation of a tangent line, one typically needs the slope and a point on the curve.
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Slopes of Tangent Lines
Derivative
The derivative of a function at a point gives the slope of the tangent line to the curve at that point. For the function y = √x, the derivative can be calculated using the power rule, which helps determine where the slope equals the given value (1/4 in this case). This is essential for finding the specific point of tangency.
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Point-Slope Form
The point-slope form of a linear equation is given by y - y₁ = m(x - x₁), where m is the slope and (x₁, y₁) is a point on the line. This form is particularly useful for writing the equation of a tangent line once the slope and the point of tangency are known. It allows for a straightforward construction of the line's equation.
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Guided course
3:56Slope-Intercept Form
Related Practice
Textbook Question
Find dy/dt when x = 1 if y = x² + 7x − 5 and dx/dt = ¹/₃.
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Textbook Question
Theory and Examples
Intersecting normal line The line that is normal to the curve x² + 2xy – 3y² = 0 at (1,1) intersects the curve at what other point?
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Textbook Question
Are there any points on the curve y = x - 1/(2x) where the slope is 2? If so, find them.
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If r + s² + v³ = 12, dr/dt = 4, and ds/dt = –3, find dv/dt when r = 3 and s = 1.
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Textbook Question
Power Rule for negative integers Use the Derivative Quotient Rule to prove the Power Rule for negative integers, that is,
d/dx (x⁻ᵐ) = −mx⁻ᵐ⁻¹
where m is a positive integer.
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