Đáp án:
$1)18x-7\\ 2)y-\dfrac{5}{\sqrt{9-2x}}\\ 4)\dfrac{36}{(7-4x)^2}\\ 9)-4\sin 8x$
Giải thích các bước giải:
$1)\\ y=9x^2-7x+3\\ y'=9.2x-7=18x-7\\ 2)\\ y=5\sqrt{9-2x}\\ y'=5.\dfrac{(9-2x)'}{2\sqrt{9-2x}}\\ =5.\dfrac{-2}{2\sqrt{9-2x}}\\ =-\dfrac{5}{\sqrt{9-2x}}\\ 4)\\ y=\dfrac{9}{7-4x}=9.\dfrac{1}{7-4x}\\ y'=9.\dfrac{-(7-4x)'}{(7-4x)^2}\\ =9.\dfrac{4}{(7-4x)^2}\\ =\dfrac{36}{(7-4x)^2}\\ 9)\\ y=\cos^24x\\ y'=2\cos 4x.(\cos 4x)'\\ =2\cos 4x.(-\sin 4x).(4x)'\\ =-8\cos 4x.\sin 4x\\ =-4\sin 8x$
