/*! 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 46 Find \(y^{\prime \prime}\) for t... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 46

Find \(y^{\prime \prime}\) for the following functions. $$y=\tan x$$

Short Answer

Expert verified
Answer: The second derivative of the function \(y = \tan x\) is \(y^{\prime \prime} = 2\sec^2x\tan x\).

Step by step solution

01

Find the first derivative, \(y^{\prime}\)

To find the first derivative of the function, \(y = \tan x\), we will use the following rule: $$(\tan x)^{\prime} = \sec^2x$$ So, the first derivative of the function y is: $$y^{\prime} = \sec^2x$$
02

Find the second derivative, \(y^{\prime \prime}\)

Now, we will find the second derivative of the function by finding the derivative of the first derivative, \(y^{\prime} = \sec^2x\), with respect to \(x\). To do this, we will use the chain rule, considering that: $$(\sec x)^{\prime} = \sec x \cdot \tan x$$ So, the second derivative of the function y is: $$y^{\prime \prime} = \frac{d}{dx}(\sec^2x) = 2\sec x \cdot \sec x \cdot \tan x = \boxed{2\sec^2x\tan x}$$

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

Graphing \(f\) and \(f^{\prime}\) a. Graph \(f\) with a graphing utility. b. Compute and graph \(f^{\prime}\) c. Verify that the zeros of \(f^{\prime}\) correspond to points at which \(f\) has \(a\) horizontal tangent line. $$f(x)=(x-1) \sin ^{-1} x \text { on }[-1,1]$$

Consider the following functions (on the given interval, if specified). Find the inverse function, express it as a function of \(x,\) and find the derivative of the inverse function. $$f(x)=3 x-4$$

Find \(f^{\prime}(x), f^{\prime \prime}(x),\) and \(f^{\prime \prime \prime}(x)\) \(f(x)=\frac{1}{x}\)

An airliner passes over an airport at noon traveling \(500 \mathrm{mi} / \mathrm{hr}\) due west. At \(1: 00 \mathrm{p} . \mathrm{m} .,\) another airliner passes over the same airport at the same elevation traveling due north at \(550 \mathrm{mi} / \mathrm{hr} .\) Assuming both airliners maintain their (equal) elevations, how fast is the distance between them changing at 2: 30 p.m.?

Use any method to evaluate the derivative of the following functions. $$f(x)=4 x^{2}-\frac{2 x}{5 x+1}$$
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Statistics

Read Explanation

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation

Geometry

Read Explanation

Pure 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.