Textbook Question
Each of Exercises 25–36 gives a formula for a function y=f(x). In each case, find f^(-1)(x) and identify the domain and range of f^(-1). As a check, show that f(f^(-1)(x))=f^(-1)(f(x))=x.
f(x) = x³ + 1
Ch. 7 - Transcendental Functions
Hass15th EditionThomas' CalculusISBN: 9780137616077
Not the one you use?Change textbookSelect a different textbook
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 7.6.37
In Exercises 21–48, find the derivative of y with respect to the appropriate variable.
37. y=s√(1-s²) + arccos(s)
Verified step by step guidance1
Identify the function to differentiate: \(y = s \sqrt{1 - s^2} + \arccos(s)\), where \(y\) is expressed in terms of \(s\).
Rewrite the square root term for easier differentiation: \(s \sqrt{1 - s^2} = s (1 - s^2)^{1/2}\).
Apply the product rule to differentiate \(s (1 - s^2)^{1/2}\). Recall the product rule: \(\frac{d}{ds}[u v] = u' v + u v'\), where \(u = s\) and \(v = (1 - s^2)^{1/2}\).
Differentiate \(\arccos(s)\) using the chain rule. The derivative of \(\arccos(s)\) with respect to \(s\) is \(-\frac{1}{\sqrt{1 - s^2}}\).
Combine the derivatives from the product rule and the derivative of \(\arccos(s)\) to write the full expression for \(\frac{dy}{ds}\).
Verified video answer for a similar problem:
This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
6mWas this helpful?
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Derivative of Composite Functions
This involves applying the chain rule when differentiating functions composed of other functions, such as √(1 - s²). The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.
Recommended video:
3:48
Evaluate Composite Functions - Special Cases
Derivative of Inverse Trigonometric Functions
The derivative of arccos(s) with respect to s is -1/√(1 - s²). Understanding the derivatives of inverse trig functions is essential for differentiating expressions involving arccos or arcsin.
Recommended video:
06:35Derivatives of Other Inverse Trigonometric Functions
Product Rule for Differentiation
When differentiating a product of two functions, such as s and √(1 - s²), the product rule is used. It states that the derivative of f(s)g(s) is f'(s)g(s) + f(s)g'(s), combining the derivatives of each function appropriately.
Recommended video:
05:18The Product Rule
Related Practice
Textbook Question
Use l’Hôpital’s rule to find the limits in Exercises 7–52.
39. lim (x → ∞) (ln 2x - ln(x + 1))
Textbook Question
Evaluate the integrals in Exercises 39–56.
52. ∫(from π/4 to π/2)cot(t)dt
1
views
Textbook Question
Each of Exercises 25–36 gives a formula for a function y=f(x). In each case, find f^(-1)(x) and identify the domain and range of f^(-1). As a check, show that f(f^(-1)(x))=f^(-1)(f(x))=x.
f(x) = (x + 3) / (x − 2)
Textbook Question
Use the results of Exercise 55 to show that the functions in Exercises 56–60 have inverses over their domains. Find a formula for df⁻¹/dx using Theorem 1.
f(x) = (1 − x)³
Textbook Question
130. Use the identity arccot(u)=π/2 - arctan(u) to derive the formula for the derivative of arccot(u) in Table 7.4 from the formula for the derivative of arctan(u).