Đạo hàm, giúp gấp với ạ

Đáp án:

$1)6\tan^2x(\tan^2x+1)+2\cot x (\cot^2x+1)\\ 2)-4\sin 4x \tan x+\cos 4x(\tan^2x+1)\\ 3)2(\tan^22x+1)\sin 5x+5\tan 2x \cos 5x\\ 4)-\dfrac{4\left(\tan^2\dfrac{4}{x+1}+1 \right)}{(x+1)^2}\\ 5)-\dfrac{\cot^2\sqrt{x+5}+1}{2\sqrt{x+5}}.$

Giải thích các bước giải:

$1)\\ y=2\tan^3x-\cot^2x\\ y'=6\tan^2x.(\tan x)'-2\cot x. (\cot x)'\\ =6\tan^2x(\tan^2x+1)+2\cot x (\cot^2x+1)\\ 2)\\ y=\cos 4x \tan x\\ y'=(\cos 4x)' \tan x+\cos 4x( \tan x)'\\ =-\sin 4x.(4x)'. \tan x+\cos 4x(\tan^2x+1)\\ =-4\sin 4x \tan x+\cos 4x(\tan^2x+1)\\ 3)\\ y=\tan 2x \sin 5x\\ y'=(\tan 2x)' \sin 5x+\tan 2x( \sin 5x)'\\ =(\tan^22x+1).(2x)' .\sin 5x+\tan 2x. \cos 5x.(5x)'\\ =2(\tan^22x+1)\sin 5x+5\tan 2x \cos 5x\\ 4)\\ y=\tan\dfrac{4}{x+1}\\ y'=\left(\tan^2\dfrac{4}{x+1}+1 \right). \left(\dfrac{4}{x+1}\right)'\\ =\left(\tan^2\dfrac{4}{x+1}+1 \right). \dfrac{-4}{(x+1)^2}.(x+1)'\\ =-\dfrac{4\left(\tan^2\dfrac{4}{x+1}+1 \right)}{(x+1)^2}\\ 5)\\ y=\cot \sqrt{x+5}\\ y'=(\cot^2\sqrt{x+5}+1).(\sqrt{x+5})'\\ =-(\cot^2\sqrt{x+5}+1).\dfrac{(x+5)'}{2\sqrt{x+5}}\\ =-\dfrac{\cot^2\sqrt{x+5}+1}{2\sqrt{x+5}}.$