교과서 질문
Derivatives in Differential Form
In Exercises 17–28, find dy.
y = (2√x)/(3(1 + √x))
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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 3.7.54
In Exercises 53 and 54, find both dy/dx (treating y as a differentiable function of x) and dx/dy (treating x as a differentiable function of y). How do dy/dx and dx/dy seem to be related?
54. x³ + y² = sin²y
검증된 단계별 안내1
Start by differentiating the given equation \(x^3 + y^2 = \sin^2(y)\) with respect to \(x\). Remember to apply implicit differentiation since \(y\) is a function of \(x\).
Differentiate \(x^3\) with respect to \(x\) to get \(3x^2\).
Differentiate \(y^2\) with respect to \(x\) using the chain rule: \(2y \cdot \frac{dy}{dx}\).
Differentiate \(\sin^2(y)\) with respect to \(x\) using the chain rule: \(2\sin(y)\cos(y) \cdot \frac{dy}{dx}\).
Set up the equation from the derivatives: \(3x^2 + 2y \cdot \frac{dy}{dx} = 2\sin(y)\cos(y) \cdot \frac{dy}{dx}\). Solve for \(\frac{dy}{dx}\) and then find \(\frac{dx}{dy}\) by taking the reciprocal of \(\frac{dy}{dx}\).
비슷한 문제에 대한 검증된 영상 답변:
이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
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주요 개념
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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 the equation x³ + y² = sin²y, both x and y are mixed together, requiring implicit differentiation to find dy/dx by differentiating both sides with respect to x, treating y as a function of x.
추천 영상:
가이드 코스
05:14
Finding The Implicit Derivative
Inverse Function Theorem
The inverse function theorem provides a relationship between the derivatives of inverse functions. If dy/dx is the derivative of y with respect to x, then dx/dy is the reciprocal of dy/dx, assuming both derivatives exist and are non-zero. This theorem helps understand how dy/dx and dx/dy are related, as they are inverses of each other.
추천 영상:
가이드 코스
06:11Fundamental Theorem of Calculus Part 1
Chain Rule
The chain rule is essential for differentiating composite functions. When finding dy/dx or dx/dy, the chain rule allows us to differentiate expressions involving y as a function of x or x as a function of y. For example, differentiating sin²y with respect to x involves using the chain rule to account for y being a function of x.
추천 영상:
05:02Intro to the Chain Rule
관련 실천
교과서 질문
[Technology Exercise]
Graph the curves in Exercises 39–48.
a. Where do the graphs appear to have vertical tangent lines?
b. Confirm your findings in part (a) with limit calculations. But before you do, read the introduction to Exercises 37 and 38.
y = 4x²/⁵ − 2x
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교과서 질문
Suppose that the function v in the Derivative Product Rule has a constant value c. What does the Derivative Product Rule then say? What does this say about the Derivative Constant Multiple Rule?
235
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교과서 질문
Derivative Calculations
In Exercises 1–12, find the first and second derivatives.
y = x² + x + 8
177
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교과서 질문
Finding Linearizations
In Exercises 1–5, find the linearization L(x) of f(x) at x = a.
f(x) = ∛x, a = −8
204
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교과서 질문
Derivatives in Differential Form
In Exercises 17–28, find dy.
xy² − 4x³/² − y = 0
186
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