Skip to main content
Ch. 3 - Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 3 - DerivativesProblem 3.7.30
Chapter 3, Problem 3.7.30

In Exercises 29 and 30, find the slope of the curve at the given points.


(x² + y²)² = (x – y)² at (1,0) and (1,–1)

Verified step by step guidance
1
First, recognize that the given equation \((x^2 + y^2)^2 = (x - y)^2\) is an implicit function of \(x\) and \(y\). To find the slope of the curve at a given point, we need to find \(\frac{dy}{dx}\) using implicit differentiation.
Differentiate both sides of the equation with respect to \(x\). For the left side, use the chain rule: \(\frac{d}{dx}((x^2 + y^2)^2) = 2(x^2 + y^2) \cdot (2x + 2y \frac{dy}{dx})\). For the right side, use the chain rule: \(\frac{d}{dx}((x - y)^2) = 2(x - y)(1 - \frac{dy}{dx})\).
Set the derivatives equal to each other: \(2(x^2 + y^2)(2x + 2y \frac{dy}{dx}) = 2(x - y)(1 - \frac{dy}{dx})\).
Simplify the equation and solve for \(\frac{dy}{dx}\). This involves expanding both sides, collecting like terms, and isolating \(\frac{dy}{dx}\) on one side of the equation.
Substitute the given points \((1, 0)\) and \((1, -1)\) into the expression for \(\frac{dy}{dx}\) to find the slope of the curve at these points. Evaluate the expression to find the numerical value of the slope at each point.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
6m
Was this helpful?

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 when it is not explicitly solved for one variable in terms of another. In this problem, the equation (x² + y²)² = (x – y)² involves both x and y, requiring implicit differentiation to find dy/dx, the slope of the curve at given points.
Recommended video:
05:14
Finding The Implicit Derivative

Chain Rule

The chain rule is a fundamental principle in calculus used to differentiate composite functions. When applying implicit differentiation to the given equation, the chain rule helps differentiate terms like (x² + y²)², where the outer function is raised to a power and the inner function involves both x and y.
Recommended video:
05:02
Intro to the Chain Rule

Evaluating Derivatives at Specific Points

After finding the derivative using implicit differentiation, it is crucial to evaluate it at specific points to determine the slope of the curve at those points. For this problem, once dy/dx is obtained, substitute the coordinates (1,0) and (1,–1) into the derivative to find the respective slopes at these points.
Recommended video:
04:50
Critical Points
Related Practice
Textbook Question

Rates of Change


Speed of a rocket At t sec after liftoff, the height of a rocket is 3t² ft. How fast is the rocket climbing 10 sec after liftoff?

204
views
Textbook Question

Find the derivatives of the functions in Exercises 19–40.


y = (5 − 2x)⁻³ + (1 / 8)(2 / x + 1)⁴

139
views
Textbook Question

If y = x² and dx/dt = 3, then what is dy/dt when x = –1?

226
views
Textbook Question

Derivative Calculations


In Exercises 1–12, find the first and second derivatives.


w = 3z⁷ − 7z³ + 21z²

247
views
Textbook Question

Approximation Error


In Exercises 29–34, each function f(x) changes value when x changes from x₀ to x₀ + dx. Find


a. the change Δf = f(x₀ + dx) − f(x₀);

b. the value of the estimate df = fʹ(x₀) dx; and

c. the approximation error |Δf − df|.


<IMAGE>


f(x) = x² + 2x, x₀ = 1, dx = 0.1

191
views
Textbook Question

Derivative Calculations


In Exercises 1–12, find the first and second derivatives.


r = 12/θ − 4/θ³ + 1/θ⁴

189
views