Đạo hàm m. y=sin^3x o. $y=sin^2 3x$ p. $y=tan^2 (4x+1)$ q. $y=cos^2 (\dfrac{2x+1}{3})$

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1. m,

   $y’=3\sin^2x(\sin x)’$

   $=3\sin^2x\cos x$

   o,

   $y’=2\sin3x.(\sin3x)’$

   $=2\sin3x.\cos3x.(3x)’

   $=3.2\sin3x\cos3x$

   $=3\sin6x$

   p,

   $y’=2\tan(4x+1).[\tan(4x+1)]’$

   $=2\tan(4x+1).\dfrac{(4x+1)’}{\cos^2(4x+1)}$

   $=\dfrac{8\tan(4x+1)}{\cos^2(4x+1)}$

   q,

   $y’=-2\cos\dfrac{2x+1}{3}.\sin\dfrac{2x+1}{3}.\dfrac{2}{3}$

   $=\dfrac{-2}{3}\sin\dfrac{4x+2}{3}$

   Trả lời

2. $m)\quad y = \sin^3x$

   $\to y’ = 3\sin^2x.(\sin x)’$

   $\to y’ = 3\sin^2x\cos x$

   $o)\quad y = \sin^23x$

   $\to y’ = 2\sin3x.(\sin3x)’$

   $\to y’ = 2\sin3x.3.\cos3x$

   $\to y’ = 3\sin6x$

   $p)\quad y= \tan^2(4x+1)$

   $\to y’ = 2\tan(4x+1).\left[\tan(4x+1)\right]’$

   $\to y’=2\tan(4x+1).\dfrac{(4x+1)’}{\cos^2(4x+1)}$

   $\to y’ = 2\tan(4x+1).\dfrac{4}{\cos^2(4x+1)}$

   $\to y’ =\dfrac{8\sin(4x+1)}{\cos^3(4x+1)}$

   $q)\quad y = \cos^2\left(\dfrac{2x+1}{3}\right)$

   $\to y’ = 2\cos\left(\dfrac{2x+1}{3}\right).\left[\cos\left(\dfrac{2x+1}{3}\right)\right]’$

   $\to y’ = 2.\cos\left(\dfrac{2x+1}{3}\right)\left[-\left(\dfrac{2x+1}{3}\right)\right]’.\sin\left(\dfrac{2x+1}{3}\right)$

   $\to y’ = 2\cos\left(\dfrac{2x+1}{3}\right).\left(-\dfrac23\right).\sin\left(\dfrac{2x+1}{3}\right)$

   $\to y’ = -\dfrac23\sin\left(\dfrac{4x+2}{3}\right)$

   Trả lời