Problem

Find $y^{\prime}$ and $y^{\prime \prime}$ $$ y=\s…

01:43

Problem 48 Medium Difficulty

Find $ y' $ and $ y". $
$ y = \frac {1}{(1 + \tan x)^2} $

Answer

$y^{\prime}=\frac{-2 \sec ^{2} x}{(1+\tan x)^{3}} y^{\prime \prime}=\frac{-2 \sec ^{3} x[-3 \sec x+2 \sin x(1+\tan x)]}{(1+\tan x)^{4}}$

Topics

Derivatives

Differentiation

Book Cover for Calculus: Early Tra…

Calculus: Early Transcendentals

Chapter 3

Differentiation Rules

Section 4

The Chain Rule

Discussion

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Watch More Solved Questions in Chapter 3

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Video Transcript

this problem's quite a lengthy application off the caution rule. Wish our write it here and because I've sort of simplification is too messy out. Simply present the final answer here. You can simply start from the original function and applied this rule and knowing that the review of tensions Second square and the reveal secrets, secret tension and simply five from there.

Frank L.

Topics

Derivatives

Differentiation

Book Cover for Calculus: Early Tra…

Calculus: Early Transcendentals

Chapter 3

Differentiation Rules

Section 4

The Chain Rule

Top Calculus 1 / AB Educators

Catherine R.

Missouri State University

Heather Z.

Oregon State University

Kristen K.

University of Michigan - Ann Arbor

Michael J.

Idaho State University