Table of contents
- 0. Functions7h 52m
- Introduction to Functions16m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms34m
- Exponential & Logarithmic Equations35m
- Introduction to Trigonometric Functions38m
- Graphs of Trigonometric Functions44m
- Trigonometric Identities47m
- Inverse Trigonometric Functions48m
- 1. Limits and Continuity2h 2m
- 2. Intro to Derivatives1h 33m
- 3. Techniques of Differentiation3h 18m
- 4. Applications of Derivatives2h 38m
- 5. Graphical Applications of Derivatives6h 2m
- 6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
- 7. Antiderivatives & Indefinite Integrals1h 26m
- 8. Definite Integrals4h 44m
- 9. Graphical Applications of Integrals2h 27m
- 10. Physics Applications of Integrals 2h 22m
4. Applications of Derivatives
Implicit Differentiation
Problem 3.8.60a
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

1
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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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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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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