Textbook Question
In Exercises 7–38, find the derivative of y with respect to x, t, or θ, as appropriate.
35. y = ln((x²+1)^5/√(1-x))
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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 7.6.23
In Exercises 21–48, find the derivative of y with respect to the appropriate variable.
23. y=arcsin(√2t)
Verified step by step guidance1
Identify the function to differentiate: \(y = \arcsin(\sqrt{2t})\) where the variable is \(t\).
Recall the derivative formula for the inverse sine function: \(\frac{d}{dx} \arcsin(u) = \frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dx}\), where \(u\) is a function of \(x\).
Set \(u = \sqrt{2t} = (2t)^{1/2}\). Find the derivative of \(u\) with respect to \(t\): \(\frac{du}{dt} = \frac{1}{2} (2t)^{-1/2} \cdot 2\) using the chain rule.
Substitute \(u\) and \(\frac{du}{dt}\) into the derivative formula: \(\frac{dy}{dt} = \frac{1}{\sqrt{1 - (\sqrt{2t})^2}} \cdot \frac{du}{dt}\).
Simplify the expression inside the square root and the derivative to write the final derivative in terms of \(t\) (without calculating the numeric value).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Derivative of Inverse Trigonometric Functions
The derivative of arcsin(x) with respect to x is 1 divided by the square root of (1 minus x squared), i.e., d/dx[arcsin(x)] = 1/√(1 - x²). This formula is essential when differentiating functions involving arcsin.
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06:35
Derivatives of Other Inverse Trigonometric Functions
Chain Rule
The chain rule is used to differentiate composite functions. If y = f(g(t)), then dy/dt = f'(g(t)) * g'(t). Here, arcsin(√2t) is a composite function requiring the chain rule to differentiate the outer and inner functions.
Recommended video:
05:02Intro to the Chain Rule
Derivative of Square Root Functions
The derivative of √u, where u is a function of t, is (1/(2√u)) * du/dt. Recognizing that √2t can be written as (2t)^(1/2) helps in applying the power rule combined with the chain rule.
Recommended video:
07:15Root Test
Related Practice
Textbook Question
Evaluate the integrals in Exercises 111–114.
111. ∫₁^(ln x) (1 / t) dt,x > 1
Textbook Question
Evaluate the integrals in Exercises 31–78.
71. ∫(from √2/3 to 2/3)dy/(|y|√(9y²-1))
Textbook Question
130. Where does the periodic function f(x) = 2e^(sin(x/2)) take on its extreme values, and what are these values?
Textbook Question
Evaluate the integrals in Exercises 77–90.
77. ∫dx/√(-x²+4x-3)
Textbook Question
In Exercises 21–48, find the derivative of y with respect to the appropriate variable.
47. y=(arccot(x³))³