Cos x'in Türevi Nedir ?
Cos x'in türevi -sin x'tir.
(cos x)′=−sin x
dxd(cos x)=−sin x
Cos x'in Türevinin İspatı
1. Yol
f′ (x)=h→0limhf (x+h)−f (x)
(cos x)′=h→0limhcos (x+h)−cos x
cos (p+q)=cos p.cos q−sin p.sin q
(cos x)′=h→0limhcos x.cos h−sin x.sin h−cos x
(cos x)′=h→0limhcos x.cos h−cos x−sin x.sin h
(cos x)′=h→0limhcos x.(cos h−1)−sin x.sin h
(cos x)′=h→0lim [hcos x.(cos h−1)−hsin x.sin h]
(cos x)′=h→0limhcos x.(cos h−1)−h→0limhsin x.sin h
(cos x)′=cos x.h→0limhcos h−1−sin x.h→0limhsin h
t→0limtsin t=1 t→0limtcos t−1=0
(cos x)′=cos x.0−sin x.1
(cos x)′=0−sin x
(cos x)′=−sin x
2. Yol
f′ (x)=h→0limhf (x+h)−f (x)
(cos x)′=h→0limhcos (x+h)−cos x
cos p−cos q=−2.sin 2p−q.sin 2p+q
(cos x)′=h→0limh−2.sin 2x+h−x.sin 2x+h+x
(cos x)′=h→0limh−2.sin 2h.sin 22x+h
(cos x)′=h→0limh−2.sin 2h.sin 22.(x+2h)
(cos x)′=h→0limh−2.sin 2h.sin (x+2h)
(cos x)′=h→0lim21.h−sin 2h.sin (x+2h)
(cos x)′=h→0lim2h−sin 2h.sin (x+2h)
(cos x)′=h→0lim [2h−sin 2h.sin (x+2h)]
(cos x)′=h→0lim2h−sin 2h.h→0limsin (x+2h)
(cos x)′=−h→0lim2hsin 2h.h→0limsin (x+2h)
h→0 (2h=h)
(cos x)′=−h→0limhsin h.h→0limsin (x+h)
(cos x)′=−1.sin (x+0)
(cos x)′=−1.sin x
(cos x)′=−sin x
3. Yol
sin x=x−3!x3+5!x5−7!x7+9!x9−...
cos x=1−2!x2+4!x4−6!x6+8!x8−...
cos x=1−2!x2+4!x4−6!x6+8!x8−...
(cos x)′=(1−2!x2+4!x4−6!x6+8!x8−...)′
(cos x)′=(1)′−(2!x2)′+(4!x4)′−(6!x6)′+(8!x8)′−...
(cos x)′=0−2!2x+4!4x3−6!6x5+8!8x7−...
(cos x)′=−2.1!2x+4.3!4x3−6.5!6x5+8.7!8x7−...
(cos x)′=−1!x+3!x3−5!x5+7!x7−...
(cos x)′=−1x+3!x3−5!x5+7!x7−...
(cos x)′=−x+3!x3−5!x5+7!x7−...
(cos x)′=−(x−3!x3+5!x5−7!x7+...)
(cos x)′=−sin x
Soru
y=cos (5x−2)⇒y′=dxdy= ?
1. Cevap
y=cos (5x−2)
dy=d[cos (5x−2)]
dy=[cos (5x−2)]′ dx
dxdy=[cos (5x−2)]′
u=5x−2
du=d(5x−2)
du=(5x−2)′ dx
du=5 dx
dxdu=5
y=cos u
dy=d(cos u)
dy=(cos u)′ du
dy=−sin u du
dudy=−sin u
dxdu.dudy=dxdy
5.−sin u=[cos (5x−2)]′
−5.sin u=[cos (5x−2)]′
−5.sin (5x−2)=[cos (5x−2)]′
2. Cevap
(cos u)′=−u′.sin u
[cos (5x−2)]′=−(5x−2)′.sin (5x−2)
[cos (5x−2)]′=−5.sin (5x−2)
Published Date:
June 25, 2020
Updated Date:
January 25, 2025