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Ch. 3 - Derivatives
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 3 - DerivativesProblem 3.8.60a
Chapter 3, Problem 3.8.60a

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

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First, we need to find the derivative of the given equation with respect to x. The equation is x + y³ - y = 1. To find the derivative, we will use implicit differentiation.
Differentiate both sides of the equation with respect to x. Remember that y is a function of x, so when differentiating terms involving y, apply the chain rule. The derivative of x is 1, the derivative of y³ is 3y²(dy/dx), and the derivative of y is (dy/dx).
After differentiating, we get: 1 + 3y²(dy/dx) - (dy/dx) = 0. Now, solve for dy/dx to find the slope of the tangent line. Combine like terms to get: dy/dx(3y² - 1) = -1.
Solve for dy/dx by dividing both sides by (3y² - 1): dy/dx = -1 / (3y² - 1). This expression gives us the slope of the tangent line at any point (x, y) on the curve.
Substitute x = 1 into the original equation to find the corresponding y-values. Solve 1 + y³ - y = 1 to find the y-values. Use these y-values in the derivative expression to find the slopes of the tangent lines. Finally, use the point-slope form of a line, y - y₁ = m(x - x₁), to write the equations of the tangent lines.

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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 differentiate equations where the dependent and independent variables are not explicitly separated. In this case, the equation x + y³ - y = 1 involves both x and y, requiring us to differentiate both sides with respect to x while treating y as a function of x. This method allows us to find dy/dx, which is essential for determining the slope of the tangent line.
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Finding The Implicit Derivative

Tangent Line Equation

The equation of a tangent line at a given point on a curve can be expressed using the point-slope form: y - y₀ = m(x - x₀), where (x₀, y₀) is the point of tangency and m is the slope at that point. Once we find the slope using implicit differentiation, we can substitute the coordinates of the point where x = 1 into this formula to derive the equation of the tangent line.
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Equations of Tangent Lines

Finding Points on the Curve

To find the equations of the tangent lines, we first need to determine the corresponding y-values for the given x-value of 1. This involves substituting x = 1 into the original equation x + y³ - y = 1 and solving for y. The solutions will provide the points on the curve where the tangent lines will be calculated, which is crucial for completing the problem.
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Summary of Curve Sketching
Related Practice
Textbook Question

Find an equation of the line tangent to the following curves at the given value of x.

y = 4 sin x cos x; x = π/3

Textbook Question

21–30. Derivatives

a. Use limits to find the derivative function f' for the following functions f.

f(x) = 1/x+1; a = -1/2;5

Textbook Question

Derivatives using tables Let h(x)=f(g(x))h(x)=f(g(x)) and p(x)=g(f(x))p(x)=g(f(x)). Use the table to compute the following derivatives.

<IMAGE>

a. h′(3)h^{\(\prime\)}\(\left\)(3\(\right\))

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)

Textbook Question

7–14. Find the derivative the following ways:

a. Using the Product Rule (Exercises 7–10) or the Quotient Rule (Exercises 11–14). Simplify your result.

g(s) = 4s³ - 8s² +4s / 4s

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Textbook Question

62–65. {Use of Tech} Graphing f and f'

b. Compute and graph f'.

f(x)=e^−x tan^−1 x on [0,∞)