Textbook Question
Derivatives in Differential Form
In Exercises 17–28, find dy.
y = 3 csc(1 − 2√x)
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 3.85
In Exercises 83–88, find equations for the lines that are tangent, and the lines that are normal, to the curve at the given point.
xy + 2x - 5y = 2, (3, 2)
Verified step by step guidance1
First, we need to find the derivative of the given equation with respect to x. The equation is xy + 2x - 5y = 2. Use implicit differentiation to differentiate both sides of the equation with respect to x.
Differentiate the left side: For the term xy, use the product rule: d(xy)/dx = x(dy/dx) + y. For the term 2x, the derivative is 2. For the term -5y, the derivative is -5(dy/dx).
Combine the derivatives: x(dy/dx) + y + 2 - 5(dy/dx) = 0. Rearrange this to solve for dy/dx, which represents the slope of the tangent line.
Substitute the given point (3, 2) into the derivative to find the slope of the tangent line at that point. This will give you the specific slope value needed for the tangent line equation.
Use the point-slope form of a line, y - y1 = m(x - x1), where m is the slope found in the previous step and (x1, y1) is the point (3, 2), to write the equation of the tangent line. For the normal line, use the negative reciprocal of the tangent slope and the same point to find its equation.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Implicit Differentiation
Implicit differentiation is a technique used to find the derivative of a function defined implicitly by an equation involving both x and y. In this case, we differentiate the equation xy + 2x - 5y = 2 with respect to x, treating y as a function of x. This allows us to find dy/dx, which is essential for determining the slope of the tangent line at a specific point.
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Finding The Implicit Derivative
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. To find the equation of the tangent line, we use the point-slope form of a line, which requires the slope (found from the derivative) and the coordinates of the point of tangency.
Recommended video:
05:13Slopes of Tangent Lines
Normal Line
The normal line to a curve at a given point is a line that is perpendicular to the tangent line at that point. Its slope is the negative reciprocal of the slope of the tangent line. To find the equation of the normal line, we again use the point-slope form, substituting the coordinates of the point and the slope of the normal line, which is derived from the tangent line's slope.
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05:13Slopes of Tangent Lines
Related Practice
Textbook Question
Find the tangent line to the Witch of Agnesi (graphed here) at the point (2,1).
Textbook Question
In Exercises 43–50, find by implicit differentiation.
__
√xy = 1
Textbook Question
The devil’s curve (Gabriel Cramer, 1750) Find the slopes of the devil’s curve y⁴ – 4y² = x⁴ – 9x² at the four indicated points.
Textbook Question
In Exercises 41–58, find dy/dt.
y = (t⁻³/⁴ sin(t))⁴/³
Textbook Question
Implicit Differentiation
In Exercises 43–50, find by implicit differentiation.
xy + 2x + 3y = 1