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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 3.7.21
3장, 문제 3.7.21

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

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Start by differentiating both sides of the equation y² = x² + 2x with respect to x. Remember that y is a function of x, so you'll need to use implicit differentiation for the left side. The derivative of y² with respect to x is 2y(dy/dx), and the derivative of x² + 2x is 2x + 2.
Set the derivatives equal to each other: 2y(dy/dx) = 2x + 2. Solve for dy/dx by dividing both sides by 2y, giving dy/dx = (2x + 2)/(2y).
Now, to find the second derivative d²y/dx², differentiate dy/dx = (2x + 2)/(2y) with respect to x again. Use the quotient rule, which states that if you have a function u/v, its derivative is (u'v - uv')/v².
Apply the quotient rule: Let u = 2x + 2 and v = 2y. Then u' = 2 and v' = 2(dy/dx). Substitute these into the quotient rule formula: d²y/dx² = [(2)(2y) - (2x + 2)(2(dy/dx))]/(2y)².
Simplify the expression for d²y/dx². Substitute dy/dx from step 2 into the expression where needed, and simplify the algebraic expression to write the second derivative in terms of x and y only.

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주요 개념

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Implicit Differentiation

Implicit differentiation is a technique used to differentiate equations where y is not isolated on one side. Instead of solving for y explicitly, we differentiate both sides of the equation with respect to x, applying the chain rule when differentiating y. This allows us to find dy/dx in terms of x and y, which is essential for problems involving relationships between variables that are not easily separable.
추천 영상:
05:14
Finding The Implicit Derivative

First Derivative (dy/dx)

The first derivative, denoted as dy/dx, represents the rate of change of the function y with respect to x. It provides information about the slope of the tangent line to the curve at any given point. In the context of implicit differentiation, finding dy/dx allows us to understand how y changes as x changes, which is crucial for further analysis, such as finding the second derivative.
추천 영상:
07:09
The First Derivative Test: Finding Local Extrema

Second Derivative (d²y/dx²)

The second derivative, denoted as d²y/dx², measures the rate of change of the first derivative. It provides insights into the curvature of the function, indicating whether the function is concave up or concave down at a point. In the context of implicit differentiation, calculating the second derivative involves differentiating dy/dx again, which often requires applying the product and chain rules, and it is essential for understanding the behavior of the function beyond just its slope.
추천 영상:
06:02
The Second Derivative Test: Finding Local Extrema
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Derivatives in Differential Form


In Exercises 17–28, find dy.


y = 3 csc(1 − 2√x)

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In Exercises 43–50, find by implicit differentiation.

__

√xy = 1

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


ƒ(t) = 1 .

2t + 1

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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)

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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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