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Ch. 3 - Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 3, Problema 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)

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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.

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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.
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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.
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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.
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Critical Points