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Ch. 3 - Derivatives
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 3, Problem 3.8.48b

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.
x⁴-x²y+y⁴=1; (−1, 1)

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First, understand that the problem requires finding the equation of the tangent line to the curve defined by the equation x⁴ - x²y + y⁴ = 1 at the point (-1, 1). This involves using implicit differentiation to find the derivative dy/dx.
Differentiate both sides of the equation x⁴ - x²y + y⁴ = 1 with respect to x. Remember to apply the product rule to the term x²y and the chain rule to y⁴. The derivative of x⁴ is 4x³, and the derivative of y⁴ with respect to x is 4y³(dy/dx).
For the term x²y, apply the product rule: the derivative is 2xy + x²(dy/dx). Combine all these derivatives to form the equation: 4x³ - (2xy + x²(dy/dx)) + 4y³(dy/dx) = 0.
Solve the resulting equation for dy/dx, which represents the slope of the tangent line at any point (x, y) on the curve. Substitute the given point (-1, 1) into the equation to find the specific slope at that point.
Once you have the slope, use the point-slope form of the equation of a line, y - y₁ = m(x - x₁), where m is the slope and (x₁, y₁) is the given point (-1, 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 involving both x and y. In this case, the equation x⁴ - x²y + y⁴ = 1 requires us to differentiate both sides 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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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 of the tangent line. Once the 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) on the curve.
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Slope of the Tangent Line

The slope of the tangent line at a point on a curve represents the instantaneous rate of change of the function at that point. It is calculated as the derivative of the function evaluated at the specific x-coordinate. In this problem, finding the slope at the point (-1, 1) is crucial for constructing the tangent line equation, as it directly influences the line's steepness and direction.
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