Textbook Question
In Exercises 59–86, find the derivative of y with respect to the given independent variable.
75. y = x³ log₁₀ x
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 Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 7, Problem 7.6.115
Solve the initial value problems in Exercises 115–120.
115. dy/dx = 1/√(1 - x²), y(0) = 0
Verified step by step guidance1
Identify the differential equation given: \(\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}\) with the initial condition \(y(0) = 0\).
Recognize that to find \(y\), you need to integrate the right-hand side with respect to \(x\): \(y = \int \frac{1}{\sqrt{1 - x^2}} \, dx + C\).
Recall the integral formula: \(\int \frac{1}{\sqrt{1 - x^2}} \, dx = \arcsin(x) + C\).
Apply the initial condition \(y(0) = 0\) to solve for the constant of integration \(C\): substitute \(x=0\) and \(y=0\) into \(y = \arcsin(x) + C\).
Write the final solution as \(y = \arcsin(x) + C\) with the value of \(C\) found from the initial condition.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Separable Differential Equations
A separable differential equation can be written as a product of a function of x and a function of y, allowing the variables to be separated on opposite sides of the equation. This enables integration with respect to each variable independently to find the general solution.
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Solving Separable Differential Equations
Integration of Inverse Trigonometric Functions
The integral of 1/√(1 - x²) is the inverse sine function, arcsin(x), plus a constant. Recognizing this form is essential for solving differential equations involving derivatives that match inverse trigonometric derivatives.
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Initial Value Problems (IVP)
An initial value problem specifies the value of the solution at a particular point, allowing determination of the constant of integration. This ensures a unique solution that satisfies both the differential equation and the initial condition.
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Related Practice
Textbook Question
Solve the initial value problems in Exercises 55–58.
55. dy/dt = e^t sin(e^t − 2),y(ln 2) = 0
Textbook Question
Find the limits in Exercises 13–20. (If in doubt, look at the function’s graph.)
19. lim(x→∞)arccsc(x)
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Textbook Question
Use l’Hôpital’s rule to find the limits in Exercises 7–52.
9. lim (t → -3) (t³ - 4t + 15) / (t² - t - 12)
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Textbook Question
In Exercises 21–48, find the derivative of y with respect to the appropriate variable.
41. y= x arcsin(x) + √(1-x²)
Textbook Question
In Exercises 7–38, find the derivative of y with respect to x, t, or θ, as appropriate.
18. y = t√(ln t)
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