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 3h 16m
- 11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
6. Derivatives of Inverse, Exponential, & Logarithmic Functions
Derivatives of Inverse Trigonometric Functions
Problem 3.10.44
Textbook Question
Tangent lines Find an equation of the line tangent to the graph of f at the given point.
f(x) = sec−1(ex); (ln 2,π/3)

1
First, understand that the problem requires finding the equation of the tangent line to the function f(x) = sec^(-1)(e^x) at the point (ln 2, π/3). The general form of a tangent line equation is y - y1 = m(x - x1), where m is the slope of the tangent line and (x1, y1) is the given point.
To find the slope m of the tangent line, we need to compute the derivative of f(x) = sec^(-1)(e^x). Recall that the derivative of sec^(-1)(u) with respect to x is 1 / (|u| * sqrt(u^2 - 1)) * du/dx. Here, u = e^x, so we need to apply the chain rule.
Calculate the derivative of u = e^x, which is du/dx = e^x. Substitute this into the derivative formula for sec^(-1)(u) to get f'(x) = 1 / (|e^x| * sqrt((e^x)^2 - 1)) * e^x.
Evaluate the derivative at the given point x = ln 2. Substitute x = ln 2 into f'(x) to find the slope m of the tangent line. This involves calculating e^(ln 2) and simplifying the expression.
Finally, use the point-slope form of the line equation with the point (ln 2, π/3) and the slope m found in the previous step to write the equation of the tangent line. Substitute x1 = ln 2, y1 = π/3, and m into the equation y - y1 = m(x - x1).

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Tangent Line
A tangent line to a curve at a given point is a straight line that touches the curve at that point without crossing it. The slope of the tangent line is equal to the derivative of the function at that point, which represents the instantaneous rate of change of the function.
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Derivative
The derivative of a function measures how the function's output value changes as its input value changes. It is defined as the limit of the average rate of change of the function over an interval as the interval approaches zero. For the function f(x) = sec^−1(e^x), finding the derivative is essential to determine the slope of the tangent line at the specified point.
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Inverse Functions
Inverse functions reverse the effect of the original function. For example, if f(x) = sec^−1(x), then its inverse would be sec(x). Understanding how to differentiate inverse functions is crucial, as it involves applying the chain rule and recognizing the relationship between a function and its inverse when calculating derivatives.
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