Cho hàm số fx=3(x+1).cos Tính f'(x),f”(x) 25/10/2021 Bởi Clara Cho hàm số fx=3(x+1).cos Tính f'(x),f”(x)
$f(x)=3(x+1)cos u$ $f'(x)=3(x+1)’\cos u+3(x+1)(\cos u)’=3\cos u-3u(x+1)\sin u$ $f”(x)=-3\sin u-[3u'(x+1)\sin u+3u(x+1)’\sin u+3u(x+1)(\sin u’)]$ $=-3\sin u-3u'(x+1)\sin u+3u\sin u+3u^2(x+1)\cos u$ Bình luận
Ta có $f'(x) = 3 (x+1)’ \cos x + 3 (x+1) (\cos x)’$ $= 3\cos x – 3 (x+1) \sin x$ Suy ra $f”(x) = [f'(x)]’$ $= 3 (\cos x)’ – 3 [(x+1) \sin x]’$ $= -3\sin x – 3[(x+1)’ \sin x + (x+1)(\sin x)’]$ $= -3\sin x – 3[\sin x + (x+1)\cos x]$ $= -6\sin x – 3(x+1)\cos x$ Vậy $f'(x) = 3\cos x – 3 (x+1) \sin x$, $f”(x) =-6\sin x – 3(x+1)\cos x$. Bình luận
$f(x)=3(x+1)cos u$
$f'(x)=3(x+1)’\cos u+3(x+1)(\cos u)’=3\cos u-3u(x+1)\sin u$
$f”(x)=-3\sin u-[3u'(x+1)\sin u+3u(x+1)’\sin u+3u(x+1)(\sin u’)]$
$=-3\sin u-3u'(x+1)\sin u+3u\sin u+3u^2(x+1)\cos u$
Ta có
$f'(x) = 3 (x+1)’ \cos x + 3 (x+1) (\cos x)’$
$= 3\cos x – 3 (x+1) \sin x$
Suy ra
$f”(x) = [f'(x)]’$
$= 3 (\cos x)’ – 3 [(x+1) \sin x]’$
$= -3\sin x – 3[(x+1)’ \sin x + (x+1)(\sin x)’]$
$= -3\sin x – 3[\sin x + (x+1)\cos x]$
$= -6\sin x – 3(x+1)\cos x$
Vậy
$f'(x) = 3\cos x – 3 (x+1) \sin x$, $f”(x) =-6\sin x – 3(x+1)\cos x$.