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Ch. 4 - Applications of the Derivative
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 4 - Applications of the DerivativeProblem 3.8.13a
Chapter 4, Problem 3.8.13a

13-26 Implicit differentiation Carry out the following steps.
a. Use implicit differentiation to find dy/dx.
x⁴+y⁴ = 2;(1,−1)

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Start by differentiating both sides of the equation \(x^4 + y^4 = 2\) with respect to \(x\). Remember that \(y\) is a function of \(x\), so you'll need to use implicit differentiation for terms involving \(y\).
Differentiate \(x^4\) with respect to \(x\) to get \(4x^3\).
Differentiate \(y^4\) with respect to \(x\) using the chain rule. This gives \(4y^3 \frac{dy}{dx}\).
Set the derivative of the left side equal to the derivative of the right side, which is zero, since the derivative of a constant is zero: \(4x^3 + 4y^3 \frac{dy}{dx} = 0\).
Solve for \(\frac{dy}{dx}\) by isolating it on one side of the equation: \(\frac{dy}{dx} = -\frac{x^3}{y^3}\). Finally, substitute the point \((1, -1)\) into the expression to find the specific value of \(\frac{dy}{dx}\) at that point.

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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. Instead of solving for y in terms of x, we differentiate both sides of the equation with respect to x, treating y as a function of x. This method allows us to find the derivative dy/dx without isolating y, which is particularly useful for complex equations.
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Finding The Implicit Derivative

Chain Rule

The chain rule is a fundamental principle in calculus that allows us to differentiate composite functions. When using implicit differentiation, we apply the chain rule to account for the derivative of y with respect to x, denoted as dy/dx. This means that when differentiating terms involving y, we multiply by dy/dx to correctly represent the relationship between the variables.
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Intro to the Chain Rule

Evaluating Derivatives at a Point

After finding the expression for dy/dx through implicit differentiation, we often need to evaluate this derivative at a specific point, such as (1, -1) in this case. This involves substituting the x and y values into the derived expression to find the slope of the tangent line at that point. This step is crucial for understanding the behavior of the function at specific coordinates.
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Related Practice
Textbook Question

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Textbook Question

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Textbook Question

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Textbook Question

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Textbook Question

Sketch the graph of a function continuous on the given interval that satisfies the following conditions.

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