Textbook Question
In Exercises 59–86, find the derivative of y with respect to the given independent variable.
85. y = log₂(8t^(ln 2))
Ch. 7 - Transcendental Functions
Hass15th EditionThomas' CalculusISBN: 9780137616077
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Table of contents
- 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
Chapter 7, Problem 7.3.43
Evaluate the integrals in Exercises 33–54.
∫ 2t e^(-t²) dt
Verified step by step guidance1
Identify the integral to solve: \(\int 2t e^{-t^{2}} \, dt\).
Recognize that the integrand contains a function and its derivative: the exponent \(-t^{2}\) and the factor \$2t$ which is the derivative of $-t^{2}$ up to a constant.
Use substitution by letting \(u = -t^{2}\). Then compute the differential: \(du = -2t \, dt\), which implies \(-du = 2t \, dt\).
Rewrite the integral in terms of \(u\): \(\int 2t e^{-t^{2}} \, dt = \int e^{u} (-du) = -\int e^{u} \, du\).
Integrate with respect to \(u\): \(-\int e^{u} \, du = -e^{u} + C\), then substitute back \(u = -t^{2}\) to get the final expression in terms of \(t\).
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1mKey Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Integration by Substitution
Integration by substitution is a method used to simplify integrals by changing variables. It involves identifying a part of the integrand as a new variable, which transforms the integral into a simpler form. This technique is especially useful when the integral contains a function and its derivative.
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Substitution With an Extra Variable
Exponential Functions in Integration
Exponential functions, such as e^(x), often appear in integrals and require special attention. Understanding how to integrate expressions involving e raised to a function is crucial, especially when combined with polynomial terms. Recognizing patterns helps in applying substitution effectively.
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Differentiation of Composite Functions
Differentiation of composite functions involves applying the chain rule to find derivatives of functions within functions. This concept is important in integration when reversing the differentiation process, as it helps identify suitable substitutions by recognizing inner functions and their derivatives.
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Related Practice
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