Ch. 3 - Derivatives

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 3 - Derivatives

Problem 3.8.10

Chapter 3, Problem 3.8.10

Find the slope of the curve x²+y³=2 at each point where y=1 (see figure). <IMAGE>

Verified step by step guidance

First, understand that the problem requires finding the slope of the curve at points where y = 1. The slope of a curve at a point is given by the derivative of the curve with respect to x, which is dy/dx.

To find dy/dx, we need to differentiate the given equation implicitly. The equation is x² + y³ = 2. Differentiate both sides with respect to x. Remember that y is a function of x, so when differentiating y³, use the chain rule.

Differentiating x² with respect to x gives 2x. Differentiating y³ with respect to x gives 3y²(dy/dx) using the chain rule. The derivative of the constant 2 is 0.

Set up the equation from the differentiation: 2x + 3y²(dy/dx) = 0. Solve for dy/dx to find the slope of the curve. Rearrange the equation to isolate dy/dx: dy/dx = -2x / 3y².

Substitute y = 1 into the equation dy/dx = -2x / 3y² to find the slope at points where y = 1. This simplifies to dy/dx = -2x / 3. The slope at these points depends on the value of x, which can be found by substituting y = 1 back into the original equation to solve for x.

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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. Instead of solving for y explicitly, we differentiate both sides of the equation with respect to x, treating y as a function of x. This allows us to find dy/dx, which represents the slope of the curve at any given point.

Slope of a Curve

The slope of a curve at a given point is defined as the rate of change of the y-coordinate with respect to the x-coordinate at that point. Mathematically, it is represented by the derivative dy/dx. For a curve defined by an equation, the slope can be evaluated by substituting the coordinates of the point into the derivative obtained through implicit differentiation.

Evaluating Derivatives at Specific Points

Once the derivative of the curve is found, evaluating it at specific points involves substituting the x and y values of those points into the derivative expression. In this case, since we are interested in points where y=1, we will first find the corresponding x values from the original equation and then substitute these into the derivative to find the slope at those points.

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Textbook Question

Find the derivative of the following functions.

y = e^-x sin x

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27–76. Calculate the derivative of the following functions.

$\sqrt[3]{x^{2} + 9}$

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Derivatives of products and quotients Find the derivative of the following functions by first expanding or simplifying the expression. Simplify your answers.

h(x) = √x (√x-x³/²)

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13–40. Evaluate the derivative of the following functions.

f(x) = 1/tan^−1(x²+4)

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Textbook Question

5–8. Calculate dy/dx using implicit differentiation.

x = y²

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