Tính đạo hàm y = tan ³ 2x – cos ( 3x + π/5) ² 11/11/2021 Bởi Adeline Tính đạo hàm y = tan ³ 2x – cos ( 3x + π/5) ²
$y’=$$[tan^3 2x – cos^2 ( 3x + π/5)]’$ = $3.(tan^2 2x).(tan 2x)’ – 2.cos ( 3x + π/5).cos ( 3x + π/5)’$ = $3.(tan^2 2x).\dfrac{2}{cos^22x} + 2.cos ( 3x + π/5).sin(3x + π/5).3$ = $3.(tan^2 2x).\dfrac{2}{cos^22x} + 6.cos ( 3x + π/5).sin(3x + π/5)$ Bình luận
$y’=3\tan^22x.(\tan2x)’+\sin(3x+\dfrac{\pi}{5})[(3x+\dfrac{\pi}{5})^2]’$ $=6\tan^22x.\dfrac{1}{\cos^22x}+\sin(3x+\dfrac{\pi}{5}.2(3x+\dfrac{\pi}{5})(3x+\dfrac{\pi}{5})’$ $=\dfrac{6\tan^22x}{\cos^22x}+6\sin(3x+\dfrac{\pi}{5})(3x+\dfrac{\pi}{5})$ Bình luận
$y’=$$[tan^3 2x – cos^2 ( 3x + π/5)]’$
= $3.(tan^2 2x).(tan 2x)’ – 2.cos ( 3x + π/5).cos ( 3x + π/5)’$
= $3.(tan^2 2x).\dfrac{2}{cos^22x} + 2.cos ( 3x + π/5).sin(3x + π/5).3$
= $3.(tan^2 2x).\dfrac{2}{cos^22x} + 6.cos ( 3x + π/5).sin(3x + π/5)$
$y’=3\tan^22x.(\tan2x)’+\sin(3x+\dfrac{\pi}{5})[(3x+\dfrac{\pi}{5})^2]’$
$=6\tan^22x.\dfrac{1}{\cos^22x}+\sin(3x+\dfrac{\pi}{5}.2(3x+\dfrac{\pi}{5})(3x+\dfrac{\pi}{5})’$
$=\dfrac{6\tan^22x}{\cos^22x}+6\sin(3x+\dfrac{\pi}{5})(3x+\dfrac{\pi}{5})$