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Ch. 3 - Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077당신이 사용하는 게 아니라요?교과서 변경
모든 교과서Hass 15th EditionCh. 3 - Derivatives문제 3.7.29
3장, 문제 3.7.29

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


y² + x² = y⁴ – 2x at (–2,1) and (–2,–1)

검증된 단계별 안내
1
First, recognize that the given equation y² + x² = y⁴ – 2x is an implicit function. To find the slope of the curve at a given point, we need to differentiate both sides of the equation with respect to x using implicit differentiation.
Differentiate the left side of the equation: The derivative of y² with respect to x is 2y(dy/dx) and the derivative of x² is 2x.
Differentiate the right side of the equation: The derivative of y⁴ with respect to x is 4y³(dy/dx) and the derivative of -2x is -2.
Set up the equation from the derivatives: 2y(dy/dx) + 2x = 4y³(dy/dx) - 2. Rearrange this equation to solve for dy/dx, which represents the slope of the curve.
Substitute the given points (-2,1) and (-2,-1) into the equation for dy/dx to find the slope at each point. This involves substituting x = -2 and y = 1 for the first point, and x = -2 and y = -1 for the second point, then solving for dy/dx in each case.

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이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
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3m

주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Implicit Differentiation

Implicit differentiation is a technique used to find the derivative of functions that are not explicitly solved for one variable in terms of another. In equations like y² + x² = y⁴ – 2x, where y is not isolated, implicit differentiation allows us to differentiate both sides with respect to x, treating y as a function of x, and applying the chain rule where necessary.
추천 영상:
가이드 코스
05:14
Finding The Implicit Derivative

Slope of a Curve

The slope of a curve at a given point is the derivative of the function at that point, representing the rate of change of the function with respect to x. For a curve defined implicitly, the slope can be found by differentiating the equation and solving for dy/dx, which gives the slope of the tangent line to the curve at the specified point.
추천 영상:
가이드 코스
05:13
Slopes of Tangent Lines

Substitution of Points

Substitution involves plugging specific coordinates into the derivative to find the slope at those points. After finding the general expression for dy/dx using implicit differentiation, substitute the given points, such as (–2,1) and (–2,–1), into this expression to calculate the slope of the curve at each point, ensuring the points satisfy the original equation.
추천 영상:
04:27
Substitution With an Extra Variable
관련 실천
교과서 질문

Slopes on the graph of the tangent function Graph y = tan x and its derivative together on (−π/2, π/2). Does the graph of the tangent function appear to have a smallest slope? A largest slope? Is the slope ever negative? Give reasons for your answers.

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교과서 질문

Derivatives in Differential Form


In Exercises 17–28, find dy.


y = 3 csc(1 − 2√x)

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교과서 질문

Second Derivatives


In Exercises 19–26, use implicit differentiation to find dy/dx and then d²y/dx². Write the solutions in terms of x and y only.


y² = x² + 2x

357
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교과서 질문

In Exercises 65 and 66, find the derivative using the definition.


ƒ(t) = 1 .

2t + 1

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교과서 질문

For Exercises 55 and 56, evaluate each limit by first converting each to a derivative at a particular x-value.


lim (x → −1) (x²/⁹ − 1) / (x + 1)

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교과서 질문

Normal lines to a parabola Show that if it is possible to draw three normal lines from the point (a, 0) to the parabola x = y² shown in the accompanying diagram, then a must be greater than 1/2. One of the normal lines is the x-axis. For what value of a are the other two normal lines perpendicular?


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