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 3.10.64c
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)=(sec^−1 x)/x on [1,∞)

1
Step 1: Understand the function f(x) = \frac{\sec^{-1}(x)}{x} and its domain [1, \infty). The function involves the inverse secant function, which is defined for x \geq 1.
Step 2: Find the derivative f'(x) using the quotient rule. The quotient rule states that if you have a function h(x) = \frac{u(x)}{v(x)}, then h'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}. Here, u(x) = \sec^{-1}(x) and v(x) = x.
Step 3: Calculate the derivatives u'(x) and v'(x). For u(x) = \sec^{-1}(x), use the derivative formula \frac{d}{dx}[\sec^{-1}(x)] = \frac{1}{|x|\sqrt{x^2 - 1}}. For v(x) = x, the derivative v'(x) = 1.
Step 4: Substitute u'(x), u(x), v'(x), and v(x) into the quotient rule formula to find f'(x). Simplify the expression to get the derivative in a manageable form.
Step 5: Set f'(x) = 0 to find the zeros of the derivative. These zeros correspond to the x-values where the original function f(x) has horizontal tangent lines. Verify these points by checking the graph of f(x) and f'(x) using technology.

This video solution was recommended by our tutors as helpful for the problem above
Video duration:
8mPlay a video:
Was this helpful?
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'(x), represents the rate of change of the function f(x) at any point x. Critical points occur where the derivative is zero or undefined, indicating potential locations for horizontal tangent lines. Understanding how to find and interpret these points is essential for analyzing the behavior of the function.
Recommended video:
Critical Points
Horizontal Tangent Lines
A horizontal tangent line occurs at points where the derivative of a function equals zero, meaning the slope of the tangent line is flat. This indicates that the function is neither increasing nor decreasing at that point, which is crucial for identifying local maxima, minima, or points of inflection. Verifying these points involves checking where f'(x) = 0.
Recommended video:
Slopes of Tangent Lines
Graphing Functions and Their Derivatives
Graphing a function f(x) alongside its derivative f'(x) provides visual insight into the function's behavior. The zeros of f' correspond to the x-values where f has horizontal tangents, allowing for a clear understanding of how the function behaves at those points. This graphical representation aids in confirming the relationship between a function and its derivative.
Recommended video:
Graphing The Derivative
Watch next
Master Derivatives of Inverse Sine & Inverse Cosine with a bite sized video explanation from Callie
Start learning