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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 3.8.46b
Chapter 3, Problem 3.8.46b

45–50. Tangent lines Carry out the following steps. <IMAGE>
b. Determine an equation of the line tangent to the curve at the given point.
x³+y³=2xy; (1, 1)

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First, identify the given curve equation: \(x^3 + y^3 = 2xy\). We need to find the derivative to determine the slope of the tangent line at the point (1, 1).
Use implicit differentiation to differentiate both sides of the equation with respect to \(x\). Remember that \(y\) is a function of \(x\), so apply the chain rule when differentiating terms involving \(y\).
Differentiate the left side: \(\frac{d}{dx}(x^3) + \frac{d}{dx}(y^3) = 3x^2 + 3y^2 \frac{dy}{dx}\).
Differentiate the right side: \(\frac{d}{dx}(2xy) = 2y + 2x \frac{dy}{dx}\).
Set the derivatives equal: \(3x^2 + 3y^2 \frac{dy}{dx} = 2y + 2x \frac{dy}{dx}\). Solve for \(\frac{dy}{dx}\) to find the slope of the tangent line at the point (1, 1). Substitute \(x = 1\) and \(y = 1\) into the derivative to find the specific slope at that point. Finally, use the point-slope form of a line, \(y - y_1 = m(x - x_1)\), to write the equation of the tangent line.

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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, rather than explicitly as y = f(x). In this case, the equation x³ + y³ = 2xy 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 the derivative (slope) is calculated using implicit differentiation, this formula can be applied to find the specific equation of the tangent line at the point (1, 1) for the given curve.
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Equations of Tangent Lines

Slope of the Tangent Line

The slope of the tangent line represents the instantaneous rate of change of the function at a specific point. In the context of the curve defined by the equation x³ + y³ = 2xy, the slope can be found by evaluating the derivative dy/dx at the point (1, 1). This slope is crucial for constructing the tangent line, as it indicates how steep the line will be at that point on the curve.
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Related Practice
Textbook Question

Use the graph of g in the figure to do the following. <IMAGE>

b. Find the values of x in (-2,2) at which g is not differentiable.

Textbook Question

Vertical tangent lines

b. Does the curve have any horizontal tangent lines? Explain.

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Textbook Question

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.

b. d/dx(tan^−1 x) =sec² x

Textbook Question

60–62. {Use of Tech} Multiple tangent lines Complete the following steps. <IMAGE>

b. Graph the tangent lines on the given graph.

4x³ =y²(4−x); x=2 (cissoid of Diocles)

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Textbook Question

Explain why or why not. Determine whether the following statements are true and give an explanation or counterexample.


b. ln(x + 1) + ln(x − 1) = ln(x² − 1), for all x.

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Textbook Question

{Use of Tech} Computing limits with angles in degrees Suppose your graphing calculator has two functions, one called sin x, which calculates the sine of x when x is in radians, and the other called s(x), which calculates the sine of x when x is in degrees.

b. Evaluate lim x→0 s(x) / x. Verify your answer by estimating the limit on your calculator.