Ch. 3 - Derivatives

Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo

Hass 15th Edition

Ch. 3 - Derivatives

Problema 3.5.33a

Capitolo 3, Problema 3.5.33a

Find y'' if:

a. y = csc x

Guida verificata passo dopo passo

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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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.

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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.

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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.

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Finding Differentials

Pratica correlata

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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:

i. Q(a) = f(a)

ii. Q′(a) = f′(a)

iii. Q″(a) = f″(a).

Determine the coefficients b₀, b₁, and b₂.

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Interpreting Derivative Values

Effectiveness of a drug On a scale from 0 to 1, the effectiveness E of a pain-killing drug t hours after entering the bloodstream is displayed in the accompanying figure.

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

189

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Domanda del libro di testo

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.

196

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Domanda del libro di testo