Cos 2x'in Türevi

Cos 2x'in Türevi Nedir ?

cos 2x'in türevi, -2.sin2x'tir.

$\frac{d}{dx}(\cos 2x)=-2.\sin 2x$

Cos 2x'in Türevinin İspatı

1. Yol

$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}$

$(\cos 2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{\cos 2(x+h)-\cos 2x}{h}$

$(\cos 2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{\cos (2x+2h)-\cos 2x}{h}$

$\boxed{\cos (p+q)=\cos p.\cos q-\sin p.\sin q}$

$(\cos 2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{\cos 2x.\cos 2h-\sin 2x.\sin 2h-\cos 2x}{h}$

$(\cos 2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{\cos 2x.\cos 2h-\cos 2x-\sin 2x.\sin 2h}{h}$

$(\cos 2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{\cos 2x.(\cos 2h-1)-\sin 2x.\sin 2h}{h}$

$(\cos 2x{)}^{\prime }=\underset{h\to 0}{\lim }[\frac{\cos 2x.(\cos 2h-1)-\sin 2x.\sin 2h}{h}.\frac{2}{2}]$

$(\cos 2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{2.[\cos 2x.(\cos 2h-1)-\sin 2x.\sin 2h]}{2.h}$

$(\cos 2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{2.\cos 2x.(\cos 2h-1)-2.\sin 2x.\sin 2h}{2h}$

$(\cos 2x{)}^{\prime }=\underset{h\to 0}{\lim }[\frac{2.\cos 2x.(\cos 2h-1)}{2h}-\frac{2.\sin 2x.\sin 2h}{2h}]$

$(\cos 2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{2.\cos 2x.(\cos 2h-1)}{2h}-\underset{h\to 0}{\lim }\frac{2.\sin 2x.\sin 2h}{2h}$

$(\cos 2x{)}^{\prime }=2.\cos 2x.\underset{h\to 0}{\lim }\frac{\cos 2h-1}{2h}-2.\sin 2x.\underset{h\to 0}{\lim }\frac{\sin 2h}{2h}$

$h=2h$$(h\to 0)$

$(\cos 2x{)}^{\prime }=2.\cos 2x.\underset{h\to 0}{\lim }\frac{\cos h-1}{h}-2.\sin 2x.\underset{h\to 0}{\lim }\frac{\sin h}{h}$

$\boxed{\underset{t\to 0}{\lim }\frac{\sin t}{t}=1}$ $\boxed{\underset{t\to 0}{\lim }\frac{\cos t-1}{t}=0}$

$(\cos 2x{)}^{\prime }=2.\cos 2x.0-2.\sin 2x.1$

$(\cos 2x{)}^{\prime }=0-2.\sin 2x$

$(\cos 2x{)}^{\prime }=-2.\sin 2x$

2. Yol

$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}$

$(\cos 2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{\cos 2(x+h)-\cos 2x}{h}$

$(\cos 2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{\cos (2x+2h)-\cos 2x}{h}$

$\boxed{\cos p-\cos q=-2.\sin (\frac{p-q}{2}).\sin (\frac{p+q}{2})}$

$(\cos 2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{-2.\sin (\frac{\cancel{2x}+2h-\cancel{2x}}{2}).\sin (\frac{2x+2h+2x}{2})}{h}$

$(\cos 2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{-2.\sin (\frac{2h}{2}).\sin (\frac{4x+2h}{2})}{h}$

$(\cos 2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{-2.\sin (\frac{\cancel{2}.h}{\cancel{2}}).\sin [\frac{\cancel{2}.(2x+h)}{\cancel{2}}]}{h}$

$(\cos 2x{)}^{\prime }=\underset{h\to 0}{\lim }[\frac{-2.\sin h}{h}.\sin (2x+h)]$

$(\cos 2x{)}^{\prime }=\underset{h\to 0}{\lim }\frac{-2.\sin h}{h}.\underset{h\to 0}{\lim }\sin (2x+h)$

$(\cos 2x{)}^{\prime }=-2.\underset{h\to 0}{\lim }\frac{\sin h}{h}.\underset{h\to 0}{\lim }\sin (2x+h)$

$\boxed{\underset{t\to 0}{\lim }\frac{\sin t}{t}=1}$

$(\cos 2x{)}^{\prime }=-\mathrm{2.1.}\sin (2x+0)$

$(\cos 2x{)}^{\prime }=-\mathrm{2.1.}\sin 2x$

$(\cos 2x{)}^{\prime }=-2.\sin 2x$

3. Yol

sin 2x ve cos 2x fonksiyonlarının sonsuz seri şeklindeki açılımlarından faydalanarak da cos 2x'in türevinin -2.sin 2x'e eşit olduğunu ispatlayabiliriz. Sin 2x ve cos 2x fonksiyonlarının sonsuz seri şeklindeki açılımları aşağıdaki gibidir.

$\boxed{\sin 2x=2x-\frac{(2x{)}^3}{3!}+\frac{(2x{)}^5}{5!}-\frac{(2x{)}^7}{7!}+\frac{(2x{)}^9}{9!}-...}$

$\boxed{\cos 2x=1-\frac{(2x{)}^2}{2!}+\frac{(2x{)}^4}{4!}-\frac{(2x{)}^6}{6!}+\frac{(2x{)}^8}{8!}-...}$

$\cos 2x=1-\frac{(2x{)}^2}{2!}+\frac{(2x{)}^4}{4!}-\frac{(2x{)}^6}{6!}+\frac{(2x{)}^8}{8!}-...$

$(\cos 2x{)}^{\prime }=[1-\frac{(2x{)}^2}{2!}+\frac{(2x{)}^4}{4!}-\frac{(2x{)}^6}{6!}+\frac{(2x{)}^8}{8!}-...{]}^{\prime }$

$(\cos 2x{)}^{\prime }=(1{)}^{\prime }-[\frac{(2x{)}^2}{2!}{]}^{\prime }+[\frac{(2x{)}^4}{4!}{]}^{\prime }-[\frac{(2x{)}^6}{6!}{]}^{\prime }+[\frac{(2x{)}^8}{8!}{]}^{\prime }-...$

$(\cos 2x{)}^{\prime }=0-\frac{2.2x.(2x{)}^{\prime }}{2!}+\frac{4.(2x{)}^3.(2x{)}^{\prime }}{4!}-\frac{6.(2x{)}^5.(2x{)}^{\prime }}{6!}+\frac{8.(2x{)}^7.(2x{)}^{\prime }}{8!}-...$

$(\cos 2x{)}^{\prime }=-\frac{2.2x.2}{2!}+\frac{4.(2x{)}^3.2}{4!}-\frac{6.(2x{)}^5.2}{6!}+\frac{8.(2x{)}^7.2}{8!}-...$

$(\cos 2x{)}^{\prime }=-\frac{\cancel{2}.2.2x}{\cancel{2}.1!}+\frac{\cancel{4}.2.(2x{)}^3}{\cancel{4}.3!}-\frac{\cancel{6}.2.(2x{)}^5}{\cancel{6}.5!}+\frac{\cancel{8}.2.(2x{)}^7}{\cancel{8}.7!}-...$

$(\cos 2x{)}^{\prime }=-\frac{2.2x}{1}+\frac{2.(2x{)}^3}{3!}-\frac{2.(2x{)}^5}{5!}+\frac{2.(2x{)}^7}{7!}-...$

$(\cos 2x{)}^{\prime }=-2.2x+\frac{2.(2x{)}^3}{3!}-\frac{2.(2x{)}^5}{5!}+\frac{2.(2x{)}^7}{7!}-...$

$(\cos 2x{)}^{\prime }=-2.[2x-\frac{(2x{)}^3}{3!}+\frac{(2x{)}^5}{5!}-\frac{(2x{)}^7}{7!}+...]$

$(\cos 2x{)}^{\prime }=-2.\sin 2x$