Table of contents
- 0. Functions7h 52m
- Introduction to Functions16m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms34m
- Exponential & Logarithmic Equations35m
- Introduction to Trigonometric Functions38m
- Graphs of Trigonometric Functions44m
- Trigonometric Identities47m
- Inverse Trigonometric Functions48m
- 1. Limits and Continuity2h 2m
- 2. Intro to Derivatives1h 33m
- 3. Techniques of Differentiation3h 18m
- 4. Applications of Derivatives2h 38m
- 5. Graphical Applications of Derivatives6h 2m
- 6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
- 7. Antiderivatives & Indefinite Integrals1h 26m
- 8. Definite Integrals4h 44m
- 9. Graphical Applications of Integrals2h 27m
- 10. Physics Applications of Integrals 2h 22m
6. Derivatives of Inverse, Exponential, & Logarithmic Functions
Derivatives of Inverse Trigonometric Functions
Problem 3.10.63c
Textbook Question
62–65. {Use of Tech} Graphing f and f'
c. Verify that the zeros of f' correspond to points at which f has a horizontal tangent line.
f(x)=(x²−1)sin^−1 x on [−1,1]

1
Step 1: Understand the problem. We need to verify that the zeros of the derivative of the function f(x) = (x^2 - 1)sin^{-1}(x) correspond to points where the original function f(x) has a horizontal tangent line.
Step 2: Find the derivative f'(x). Use the product rule for differentiation, since f(x) is a product of two functions: u(x) = x^2 - 1 and v(x) = sin^{-1}(x). The product rule states that (uv)' = u'v + uv'.
Step 3: Differentiate u(x) and v(x) separately. For u(x) = x^2 - 1, the derivative u'(x) = 2x. For v(x) = sin^{-1}(x), the derivative v'(x) = 1/√(1-x^2).
Step 4: Apply the product rule. Substitute u(x), u'(x), v(x), and v'(x) into the product rule formula to find f'(x) = (x^2 - 1)(1/√(1-x^2)) + (2x)(sin^{-1}(x)).
Step 5: Set f'(x) = 0 to find the zeros of the derivative. Solve the equation (x^2 - 1)(1/√(1-x^2)) + (2x)(sin^{-1}(x)) = 0 to find the x-values where f'(x) is zero. These x-values correspond to points where f(x) has a horizontal tangent line.

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Derivative and Critical Points
The derivative of a function, denoted as f', represents the rate of change of the function f. Critical points occur where the derivative is zero or undefined, indicating potential locations for horizontal tangent lines. In this context, finding the zeros of f' helps identify where the slope of f is zero, which corresponds to horizontal tangents.
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Critical Points
Horizontal Tangent Lines
A horizontal tangent line occurs at points on the graph of a function where the slope is zero. This means that the derivative of the function at those points is equal to zero. Understanding this concept is crucial for verifying the relationship between the zeros of the derivative and the behavior of the original function.
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Slopes of Tangent Lines
Graphing Functions
Graphing a function involves plotting its values on a coordinate plane to visualize its behavior. By graphing both f(x) and its derivative f'(x), one can observe how the zeros of f' correspond to points where f has horizontal tangents. This visual representation aids in understanding the relationship between a function and its derivative.
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Graph of Sine and Cosine Function
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