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Ch. 3 - Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077당신이 사용하는 게 아니라요?교과서 변경
3장, 문제 3.7.62b

Computer Explorations


Use a CAS to perform the following steps in Exercises 55–62.


b. Using implicit differentiation, find a formula for the derivative dy/dx and evaluate it at the given point P.


x√(1 + 2y) + y = x², P(1,0)

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Start by differentiating both sides of the equation with respect to x. The equation is x√(1 + 2y) + y = x².
Apply the product rule to the term x√(1 + 2y). The product rule states that d(uv)/dx = u'v + uv', where u = x and v = √(1 + 2y).
Differentiate u = x to get u' = 1. For v = √(1 + 2y), use the chain rule: v' = (1/2)(1 + 2y)^(-1/2) * (2(dy/dx)).
Combine the derivatives using the product rule: d(x√(1 + 2y))/dx = 1 * √(1 + 2y) + x * (1/2)(1 + 2y)^(-1/2) * 2(dy/dx).
Differentiate the remaining terms: dy/dx for y and 2x for x². Set the derivatives equal: √(1 + 2y) + x(1 + 2y)^(-1/2)(dy/dx) + dy/dx = 2x. Solve for dy/dx and evaluate at P(1,0).

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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 a function when it is not explicitly solved for one variable in terms of another. It involves differentiating both sides of an equation with respect to a variable, typically x, while treating other variables as implicit functions of x. This method is essential when dealing with equations like x√(1 + 2y) + y = x², where y is not isolated.
추천 영상:
가이드 코스
05:14
Finding The Implicit Derivative

Chain Rule

The chain rule is a fundamental principle in calculus used to differentiate composite functions. It states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. In implicit differentiation, the chain rule is crucial when differentiating terms involving y, as y is considered a function of x.
추천 영상:
05:02
Intro to the Chain Rule

Evaluating Derivatives at a Point

Once the derivative dy/dx is found using implicit differentiation, it can be evaluated at a specific point to find the slope of the tangent line at that point. This involves substituting the coordinates of the given point, such as P(1,0), into the derivative formula. This step is important for understanding the behavior of the function at specific locations on its graph.
추천 영상:
04:50
Critical Points
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교과서 질문

Computer Explorations


Use a CAS to perform the following steps in Exercises 55–62.


b. Using implicit differentiation, find a formula for the derivative dy/dx and evaluate it at the given point P.


2y² + (xy)¹/³ = x² + 2, P(1,1)

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b. area?

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y = 37 sin[(2π/365)(x − 101)] + 25


and is graphed in the accompanying figure.


b. About how many degrees per day is the temperature increasing when it is increasing at its fastest?


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Diagonals If x, y, and z are lengths of the edges of a rectangular box, then the common length of the box’s diagonals is s = √(x² + y² + z²).

b. How is ds/dt related to dy/dt and dz/dt if x is constant?

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Generalizing the Product Rule The Derivative Product Rule gives the formula


d/dx (uv) = u (dv/dx) + (du/dx) v


for the derivative of the product uv of two differentiable functions of x.


b. What is the formula for the derivative of the product u₁u₂u₃u₄ of four differentiable functions of x?

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Area The area A of a triangle with sides of lengths a and b enclosing an angle of measure θ is

A = (1/2) ab sinθ.


b. How is dA/dt related to dθ/dt and da/dt if only b is constant?

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