/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 34 Find an equation of the line tan... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 34

Find an equation of the line tangent to the graph of \(f\) at the given point. $$f(x)=\sec ^{-1}\left(e^{x}\right) ;(\ln 2, \pi / 3)$$

Short Answer

Expert verified
The equation of the tangent line is: $$y - \frac{\pi}{3} = \frac{\sqrt{3}}{3}(x - \ln 2).$$

Step by step solution

01

Find the derivative of the function

To find the derivative of \(f(x)\), we can use the chain rule, which states that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function. Since our function consists of the inverse secant function applied to a base-\(e\) exponential function, we can write \(f(x) = \sec^{-1}(u)\), with \(u = e^x\). First, we find the derivative of \(\sec^{-1}(u)\): $$\frac{d}{du}\sec^{-1}(u) = \frac{1}{|u|\sqrt{u^2 - 1}}.$$ Now, we find the derivative of \(e^x\): $$\frac{d}{dx}e^x = e^x.$$ Applying the chain rule, we obtain the derivative of \(f(x)\): $$\frac{d}{dx}f(x) = \frac{1}{|e^x|\sqrt{e^{2x} - 1}} \cdot e^x.$$
02

Evaluate the derivative at the given point

We will now evaluate the derivative \(\frac{d}{dx}f(x)\) at the point \((\ln 2, \pi / 3)\). Since \(x = \ln 2\), we can replace \(x\) with \(\ln 2\) and simplify: $$\left.\frac{d}{dx}f(x)\right|_{x=\ln 2} = \frac{1}{|2|\sqrt{2^2 - 1}} \cdot 2 = \frac{2}{2\sqrt{3}} = \frac{\sqrt{3}}{3}.$$ Therefore, the slope of the tangent line to the graph of \(f(x)\) at the point \((\ln 2, \pi / 3)\) is \(\frac{\sqrt{3}}{3}\).
03

Find the equation of the tangent line

Using the point-slope form of a linear equation, we can now find the equation of the tangent line. The point-slope form is given by: $$y - y_1 = m(x - x_1),$$ where \((x_1, y_1)\) is the point \((\ln 2, \pi / 3)\) and \(m\) is the slope \(\frac{\sqrt{3}}{3}\). Plugging in these values, we get: $$y - \frac{\pi}{3} = \frac{\sqrt{3}}{3}(x - \ln 2).$$ This is the equation of the line tangent to the graph of \(f(x) = \sec^{-1}(e^x)\) at the point \((\ln 2, \pi / 3)\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The following limits equal the derivative of a function \(f\) at a point a. a. Find one possible \(f\) and \(a\) b. Evaluate the limit. $$\lim _{x \rightarrow \pi / 4} \frac{\cot x-1}{x-\frac{\pi}{4}}$$

Proof of \(\lim _{x \rightarrow 0} \frac{\cos x-1}{x}=0\) Use the trigonometric identity \(\cos ^{2} x+\sin ^{2} x=1\) to prove that \(\lim _{x \rightarrow 0} \frac{\cos x-1}{x}=0 .(\) Hint: Begin by multiplying the numerator and denominator by \(\cos x+1 .)\)

Given the function \(f,\) find the slope of the line tangent to the graph of \(f^{-1}\) at the specified point on the graph of $$f(x)=x^{3} ;(8,2)$$

Prove the following identities and give the values of \(x\) for which they are true. $$\cos \left(\sin ^{-1} x\right)=\sqrt{1-x^{2}}$$

The following limits equal the derivative of a function \(f\) at a point a. a. Find one possible \(f\) and \(a\) b. Evaluate the limit. $$\lim _{h \rightarrow 0} \frac{\cos \left(\frac{\pi}{6}+h\right)-\frac{\sqrt{3}}{2}}{h}$$
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Geometry

Read Explanation

Pure Maths

Read Explanation

Discrete Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.