교과서 질문
In Exercises 53 and 54, find dr/ds.
r cos 2s + sin²s = π
209
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Ch. 3 - Derivatives
Hass15th EditionThomas' CalculusISBN: 9780137616077
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TOC.tableOfContents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions490
- Ch. 8 - Techniques of Integration399
- Ch. 9 - First-Order Differential Equations66
- Ch. 10 - Infinite Sequences and Series132
- Ch. 11 - Parametric Equations and Polar Coordinates58
3장, 문제 3.89
Find the slope of the curve x³y³ + y² = x + y at the points (1, 1) and (1, -1).
검증된 단계별 안내1
To find the slope of the curve at a given point, we need to find the derivative of the curve with respect to x. The curve is given by the equation: x³y³ + y² = x + y.
Use implicit differentiation to differentiate both sides of the equation with respect to x. Remember that y is a function of x, so when differentiating terms involving y, apply the chain rule.
Differentiate the left side: For x³y³, use the product rule: d/dx(x³y³) = x³ * d/dx(y³) + y³ * d/dx(x³). For y², use the chain rule: d/dx(y²) = 2y * dy/dx.
Differentiate the right side: d/dx(x) = 1 and d/dx(y) = dy/dx.
After differentiating, solve the resulting equation for dy/dx to find the expression for the slope of the curve. Substitute the points (1, 1) and (1, -1) into this expression to find the slope at each point.
비슷한 문제에 대한 검증된 영상 답변:
이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
영상 길이:
3m주요 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 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 point.
추천 영상:
가이드 코스
05:14
Finding The Implicit Derivative
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 of the function at that point. For curves defined by implicit equations, the slope can be found using the derivative obtained through implicit differentiation.
추천 영상:
11:41Summary of Curve Sketching
Evaluating Derivatives at Specific Points
Once the derivative (dy/dx) is determined, evaluating it at specific points involves substituting the coordinates of those points into the derivative expression. This gives the slope of the curve at those specific locations. In this case, we will substitute the points (1, 1) and (1, -1) into the derived expression to find the respective slopes.
추천 영상:
04:50Critical Points
관련 실천
교과서 질문
In Exercises 41–44, determine whether the piecewise-defined function is differentiable at x = 0.
f(x) = { 2x + tan x, x ≥ 0
x², x < 0
219
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교과서 질문
Heating a plate When a circular plate of metal is heated in an oven, its radius increases at the rate of 0.01 cm/min. At what rate is the plate’s area increasing when the radius is 50 cm?
249
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교과서 질문
Using the Alternative Formula for Derivatives
Use the formula
f'(x) = lim (z → x) (f(z) − f(x)) / (z − x)
to find the derivative of the functions in Exercises 23–26.
g(x) = x / (x − 1)
243
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교과서 질문
Linearization for Approximation
In Exercises 7–12, find a linearization at a suitably chosen integer near a at which the given function and its derivative are easy to evaluate.
f(x) = ∛x, a = 8.5
194
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교과서 질문
Find the derivatives of the functions in Exercises 19–40.
y = (1 / 18)(3x − 2)⁶ + (4 − (1 / 2x²))⁻¹
263
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