Cot x'in Türevi Nedir ?
Cot x'in türevi, -(1 + cot² x)'dir.
dxd(cot x)=−(1+cot2 x)
Cot x'in Türevinin İspatı
1. Yol
f′(x)=h→0limhf(x+h)−f(x)
(cot x)′=h→0limhcot (x+h)−cot x
cot (p+q)=cot p+cot qcot p.cot q−1
(cot x)′=h→0limhcot x+cot hcot x.cot h−1−cot x
(cot x)′=h→0limhcot x+cot hcot x.cot h−1−cot x.(cot x+cot h)
(cot x)′=h→0limh.(cot x+cot h)cot x.cot h−1−cot x.(cot x+cot h)
(cot x)′=h→0limh.(cot x+cot h)cot x.cot h−1−cot2 x−cot x.cot h
(cot x)′=h→0limh.(cot x+cot h)−(1+cot2 x)
cot x=tan x1
(cot x)′=h→0limh.(cot x+tan h1)−(1+cot2 x)
(cot x)′=h→0limh.cot x+tan hh−(1+cot2 x)
(cot x)′=h→0lim(h.cot x)+h→0limtan hh−(1+cot2 x)
(cot x)′=cot x.h→0limh+h→0limtan hh−(1+cot2 x)
t→0limttan t=t→0limtan tt=1
(cot x)′=cot x.0+1−(1+cot2 x)
(cot x)′=0+1−(1+cot2 x)
(cot x)′=1−(1+cot2 x)
(cot x)′=−(1+cot2 x)
2. Yol
f′(x)=h→0limhf(x+h)−f(x)
(cot x)′=h→0limhcot (x+h)−cot x
cot x=sin xcos x
(cot x)′=h→0limhsin (x+h)cos (x+h)−sin xcos x
(cot x)′=h→0limhsin x.sin (x+h)sin x.cos (x+h)−sin (x+h).cos x
(cot x)′=h→0limh.sin x.sin (x+h)sin x.cos (x+h)−sin (x+h).cos x
sin (p−q)=sin p.cos q−sin q.cos p
(cot x)′=h→0limh.sin x.sin (x+h)sin [x−(x+h)]
(cot x)′=h→0limh.sin x.sin (x+h)sin (x−x−h)
sin (−x)=−sin x
(cot x)′=h→0limh.sin x.sin (x+h)−sin h
(cot x)′=−h→0limh.sin x.sin (x+h)sin h
(cot x)′=−h→0lim(hsin h.sin x.sin (x+h)1)
(cot x)′=−h→0limhsin h.h→0limsin x.sin (x+h)1
t→0limtsin t=1
(cot x)′=−1.sin x.sin (x+0)1
(cot x)′=−sin x.sin (x+0)1
(cot x)′=−sin x.sin x1
(cot x)′=−sin2 x1
sin2 x+cos2 x=1
(cot x)′=−(sin2 xsin2 x+cos2 x)
(cot x)′=−(sin2 xsin2 x+sin2 xcos2 x)
(cot x)′=−[1+(sin xcos x)2]
(cot x)′=−(1+cot2 x)
3. Yol
İki fonksiyonun bölümünün türevi formülünden yararlanarak da cot x'in türevinin, -(1 + cot² x)'ye eşit olduğunu ispatlayabiliriz.
f(x)=v(x)u(x)⇒f′(x)=[v(x)]2u′(x).v(x)−v′(x).u(x)
cot x=sin xcos x
(cot x)′=(sin xcos x)′
(cot x)′=sin2 x(cos x)′.sinx−(sin x)′.cos x
(sin x)′=cos x (cos x)′=−sin x
(cot x)′=sin2 x−sin x.sin x−cos x.cos x
(cot x)′=sin2 x−sin2 x−cos2 x
(cot x)′=sin2 x−(sin2 x+cos2 x)
(cot x)′=−sin2 xsin2 x+cos2 x
(cot x)′=−(sin2 xsin2 x+sin2 xcos2 x)
(cot x)′=−[1+(sin xcos x)2]
(cot x)′=−(1+cot2 x)
(Cot x)' = -(1 + Cot² x) = -1/Sin² x = -Csc² x
(cot x)′=−(1+cot2 x)
cot x=sin xcos x
(cot x)′=−[1+(sin xcos x)2]
(cot x)′=−(1+sin2 xcos2 x)
(cot x)′=−(sin2 xsin2 x+cos2 x)
sin2 x+cos2 x=1
(cot x)′=−sin2 x1
(cot x)′=−(sin x1)2
csc x=sin x1
(cot x)′=−csc2 x
⇓
f(x)=cot u(x)⇒f′(x)=−u′(x).[1+cot2 u(x)] (formülün ispatını görmek için tıklayınız)
Soru:
f(x)=cot (2x3+3x2)⇒f′(x)=?
Cevap:
f(x)=cot (2x3+3x2)
f′(x)=[cot (2x3+3x2)]′
f(x)=cot u(x)⇒f′(x)=−u′(x).[1+cot2 u(x)]
f′(x)=−(2x3+3x2)′.[1+cot2 (2x3+3x2)]
f′(x)=−(3.2.x2+2.3.x).[1+cot2 (2x3+3x2)]
f′(x)=−(6x2+6x).[1+cot2 (2x3+3x2)]