## Calculus: Early Transcendentals 8th Edition

$$y'=\frac{-2\sec^2x}{(1+\tan x)^3}$$ $$y''=\frac{6\sec^4x}{(1+\tan x)^4}-\frac{4\sec^2x\tan x}{(1+\tan x)^3}$$
$$y=\frac{1}{(1+\tan x)^2}$$ $$y=(1+\tan x)^{-2}$$ 1) Find $y'$ $$y'=\frac{dy}{dx}=\frac{d(1+\tan x)^{-2}}{d(1+\tan x)}\frac{d(1+\tan x)}{dx}$$ $$y'=-2(1+\tan x)^{-3}\times(0+\sec^2 x)$$ $$y'=-2\sec^2 x(1+\tan x)^{-3}$$ $$y'=\frac{-2\sec^2x}{(1+\tan x)^3}$$ 2) Find $y''$ $$y''=-2[\sec^2 x(1+\tan x)^{-3}]'$$ $$y''=-2[(\sec^2x)'(1+\tan x)^{-3}+\sec^2x[(1+\tan x)^{-3}]']$$ $$y''=-2[\frac{d(\sec^2 x)}{d(\sec x)}\frac{d(\sec x)}{dx}](1+\tan x)^{-3}+\sec^2x[\frac{d(1+\tan x)^{-3}}{d(1+\tan x)}\frac{d(1+\tan x)}{dx}]]$$ (Here we apply the Chain Rule for both $(\sec^2x)'$ and $[(1+\tan x)^{-3}]'$) $$y''=-2[2\sec x(\sec x\tan x)](1+\tan x)^{-3}+\sec^2 x[-3(1+\tan x)^{-4}(0+\sec^2 x)]]$$ $$y''=-2[2\sec^2 x\tan x(1+\tan x)^{-3}-3\sec^4x(1+\tan x)^{-4}]$$ $$y''=6\sec^4x(1+\tan x)^{-4}-4\sec^2 x\tan x(1+\tan x)^{-3}$$ $$y''=\frac{6\sec^4x}{(1+\tan x)^4}-\frac{4\sec^2x\tan x}{(1+\tan x)^3}$$