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Ch. 3 - Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077당신이 사용하는 게 아니라요?교과서 변경
3장, 문제 3.5.33a

Find y'' if:


a. y = csc x

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First, recall that the derivative of y = csc(x) is y' = -csc(x)cot(x). This is a standard derivative that you can find in derivative tables or by using the chain rule and trigonometric identities.
Next, to find the second derivative y'', we need to differentiate y' = -csc(x)cot(x) with respect to x. This requires using the product rule, which states that if you have a function u(x)v(x), its derivative is u'(x)v(x) + u(x)v'(x).
Apply the product rule to y' = -csc(x)cot(x). Let u(x) = -csc(x) and v(x) = cot(x). Then, u'(x) = csc(x)cot(x) and v'(x) = -csc^2(x).
Substitute these derivatives into the product rule formula: y'' = u'(x)v(x) + u(x)v'(x) = [csc(x)cot(x)]cot(x) + [-csc(x)][-csc^2(x)].
Simplify the expression: y'' = csc(x)cot^2(x) + csc^3(x). This is the second derivative of y = csc(x).

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주요 개념

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

Second Derivative

The second derivative, denoted as y'', measures the rate of change of the first derivative of a function. It provides information about the concavity of the function and can indicate points of inflection where the function changes from concave up to concave down or vice versa.
추천 영상:
06:02
The Second Derivative Test: Finding Local Extrema

Cosecant Function

The cosecant function, denoted as csc x, is the reciprocal of the sine function, defined as csc x = 1/sin x. It is important to understand its properties, including its domain and range, as well as its behavior near asymptotes where sin x equals zero.
추천 영상:
가이드 코스
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Graphs of Secant and Cosecant Functions

Differentiation Rules

Differentiation rules, such as the quotient rule and chain rule, are essential for finding derivatives of complex functions. For the cosecant function, applying the quotient rule is necessary since it can be expressed as a ratio, which simplifies the process of finding its first and second derivatives.
추천 영상:
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05:53
Finding Differentials
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Differentiability and Continuity on an Interval


Each figure in Exercises 45–50 shows the graph of a function over a closed interval D. At what domain points does the function appear to be


a. differentiable?


Give reasons for your answers.

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Quadratic approximations


a. Let Q(x) = b₀ + b₁(x − a) + b₂(x − a)² be a quadratic approximation to f(x) at x = a with these properties:


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ii. Q′(a) = f′(a)

iii. Q″(a) = f″(a).


Determine the coefficients b₀, b₁, and b₂.

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Interpreting Derivative Values


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a. At what times does the effectiveness appear to be increasing? What is true about the derivative at those times?

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Differentiability and Continuity on an Interval


Each figure in Exercises 45–50 shows the graph of a function over a closed interval D. At what domain points does the function appear to be


a. differentiable?


Give reasons for your answers.

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If the original 24 m edge length x of a cube decreases at the rate of 5 m/min, when x = 3 m at what rate does the cube’s

a. surface area change?

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Differentiability and Continuity on an Interval


Each figure in Exercises 45–50 shows the graph of a function over a closed interval D. At what domain points does the function appear to be


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Give reasons for your answers.


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