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Find $ y' $ and $ y". $$ y = \frac {1}{(1 + \tan x)^2} $
$y^{\prime}=\frac{-2 \sec ^{2} x}{(1+\tan x)^{3}} y^{\prime \prime}=\frac{-2 \sec ^{3} x[-3 \sec x+2 \sin x(1+\tan x)]}{(1+\tan x)^{4}}$
Calculus 1 / AB
Chapter 3
Differentiation Rules
Section 4
The Chain Rule
Derivatives
Differentiation
Missouri State University
Oregon State University
Harvey Mudd College
University of Nottingham
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this problem's quite a lengthy application off the caution rule. Wish our write it here and because I've sort of simplification is too messy out. Simply present the final answer here. You can simply start from the original function and applied this rule and knowing that the review of tensions Second square and the reveal secrets, secret tension and simply five from there.
In mathematics, precalculus is the study of functions (as opposed to calculu…
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Find $y^{\prime}$ if tan $^{-1}(x y)=1+x^{2} y.$
Find $\frac{d y}{d x}$ if $y=\left(\frac{\tan x}{1-\tan x}\right)^{2}$.
$y=\frac{x \tan x}{x^{2}+1}$
Find $d y / d x$.$$y=\frac{1}{\tan ^{-1} x}$$
Find $d y / d x$.$$y=(\tan x)^{-1}$$
Find $d y$$$y=\tan ^{-1}\left(e^{x^{2}}\right)$$
Find $d^{2} y / d x^{2}$$$y=x \tan \left(\frac{1}{x}\right)$$
$$y^{\prime} \text { if } \tan ^{-1}(x y)=1+x^{2} y$$
Find $d y / d x$.$$y=\tan ^{-1}\left(x^{3}\right)$$
$y^{\prime \prime}=2 y+2 \tan ^{3} x, \quad y_{p}(x)=\tan x$
