Tìm đạo hàm : a. y= tan ( x+1) + cot ( 1-x) b. y= cos ²x + cos ²2x + cos ²3x 11/11/2021 Bởi Ruby Tìm đạo hàm : a. y= tan ( x+1) + cot ( 1-x) b. y= cos ²x + cos ²2x + cos ²3x
a) Ta có $y’ = [\tan(x+1) + cot(1-x)]’$ $= \dfrac{1}{\cos^2(x+1)} – \dfrac{-1}{\sin^2(1-x)}$ $= \dfrac{1}{\cos^2(x+1)} + \dfrac{1}{\sin^2(1-x)}$ Vậy $y’ = \dfrac{1}{\cos^2(x+1)} + \dfrac{1}{\sin^2(1-x)}$ b) Ta có $y’ = 2\cos x (-\sin x) + 2\cos(2x) [-2\sin(2x)] + 2\cos(3x) [-3 \sin(3x)]$ $= -2\sin x \cos x – 4\sin(2x) \cos(2x) – 6 \sin(3x) \cos(3x)$ $= -\sin(2x) – 2\sin(4x) – 3 \sin(6x)$ Vậy $y’= -\sin(2x) – 2\sin(4x) – 3 \sin(6x)$. Bình luận
a, $y’=[\tan(x+1)]’+[\cot(1-x)]’$ $=\dfrac{(x+1)’}{\cos^2(x+1)}-\dfrac{(1-x)’}{\sin^2(1-x)}$ $=\dfrac{1}{\cos^2(x+1)}+\dfrac{1}{\sin^2(x-1)}$ b, $y’=(\cos^2x)’+(\cos^22x)’+(\cos^23x)’$ $=2\cos x(\cos x)’+2\cos 2x(\cos 2x)’+2\cos 3x.(\cos 3x)’$ $=-2\cos x.\sin x -2\cos 2x.\sin 2x.(2x)’-2\cos3x.\sin3x.(3x)’$ $=-\sin2x-2\sin4x-3\sin6x$ Bình luận
a) Ta có
$y’ = [\tan(x+1) + cot(1-x)]’$
$= \dfrac{1}{\cos^2(x+1)} – \dfrac{-1}{\sin^2(1-x)}$
$= \dfrac{1}{\cos^2(x+1)} + \dfrac{1}{\sin^2(1-x)}$
Vậy
$y’ = \dfrac{1}{\cos^2(x+1)} + \dfrac{1}{\sin^2(1-x)}$
b) Ta có
$y’ = 2\cos x (-\sin x) + 2\cos(2x) [-2\sin(2x)] + 2\cos(3x) [-3 \sin(3x)]$
$= -2\sin x \cos x – 4\sin(2x) \cos(2x) – 6 \sin(3x) \cos(3x)$
$= -\sin(2x) – 2\sin(4x) – 3 \sin(6x)$
Vậy
$y’= -\sin(2x) – 2\sin(4x) – 3 \sin(6x)$.
a,
$y’=[\tan(x+1)]’+[\cot(1-x)]’$
$=\dfrac{(x+1)’}{\cos^2(x+1)}-\dfrac{(1-x)’}{\sin^2(1-x)}$
$=\dfrac{1}{\cos^2(x+1)}+\dfrac{1}{\sin^2(x-1)}$
b,
$y’=(\cos^2x)’+(\cos^22x)’+(\cos^23x)’$
$=2\cos x(\cos x)’+2\cos 2x(\cos 2x)’+2\cos 3x.(\cos 3x)’$
$=-2\cos x.\sin x -2\cos 2x.\sin 2x.(2x)’-2\cos3x.\sin3x.(3x)’$
$=-\sin2x-2\sin4x-3\sin6x$