Sin x'in Türevi (Sinüsün Türevi)

Sin x'in Türevi Nedir ?

Sin x'in türevi, cos x'tir.

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

Sin x'in Türevinin İspatı

1. Yol

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

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

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

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

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

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

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

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

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

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

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

$(\sin x{)}^{\prime }=0+\cos x$

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

2. Yol

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

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

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

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

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

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

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

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

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

$h=\frac{h}{2}$$(h\to 0)$

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

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

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

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

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

3. Yol

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

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

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

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

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

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

$(\sin x{)}^{\prime }=1-\frac{3.x^2}{3!}+\frac{5.x^4}{5!}-\frac{7.x^6}{7!}+\frac{9.x^8}{9!}-...$

$(\sin x{)}^{\prime }=1-\frac{\cancel{3}.x^2}{\cancel{3}.2!}+\frac{\cancel{5}.x^4}{\cancel{5}.4!}-\frac{\cancel{7}.x^6}{\cancel{7}.6!}+\frac{\cancel{9}.x^8}{\cancel{9}.8!}-...$

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

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

$\boxed{\Downarrow }$

$\boxed{f(x)=\sin u(x)\Rightarrow f^{\prime }(x)=u^{\prime }(x).\cos u(x)}$ (formülün ispatını görmek için tıklayınız)

$Soru:$

$f(x)=\sin (x^3-5x^2+2x+7)\Rightarrow f^{\prime }(x)=?$

$Cevap:$

$f(x)=\sin (x^3-5x^2+2x+7)$

$f^{\prime }(x)=[\sin (x^3-5x^2+2x+7){]}^{\prime }$

$\boxed{f(x)=\sin u(x)\Rightarrow f^{\prime }(x)=u^{\prime }(x).\cos u(x)}$

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

$f^{\prime }(x)=(3.x^2-\mathrm{5.2.}x+2+0).\cos (x^3-5x^2+2x+7)$

$f^{\prime }(x)=(3x^2-10x+2).\cos (x^3-5x^2+2x+7)$

$Soru:$

$f(x)=\sin 2x+\sin x^3\Rightarrow f^{\prime }(0)=?$

$Cevap:$

$f(x)=\sin 2x+\sin x^3$

$f^{\prime }(x)=(\sin 2x+\sin x^3{)}^{\prime }$

$f^{\prime }(x)=(\sin 2x{)}^{\prime }+(\sin x^3{)}^{\prime }$

$\boxed{f(x)=\sin u(x)\Rightarrow f^{\prime }(x)=u^{\prime }(x).\cos u(x)}$

$f^{\prime }(x)=(2x{)}^{\prime }.\cos 2x+(x^3{)}^{\prime }.\cos x^3$

$f^{\prime }(x)=2.\cos 2x+3.x^2.\cos x^3$

$f^{\prime }(0)=2.\cos (2.0)+3.0^2.\cos (0^3)$

$f^{\prime }(0)=2.\cos 0+\mathrm{3.0.}\cos 0$

$f^{\prime }(0)=2.\cos 0+0$

$f^{\prime }(0)=2.\cos 0$

$\boxed{\cos 0=1}$

$f^{\prime }(0)=2.1$

$f^{\prime }(0)=2$

$Soru:$

$f(x)={\sin }^{2}x\Rightarrow \frac{f^{\prime }(x)}{f(x)}=?$

$Cevap:$

$f(x)={\sin }^{2}x$

$f^{\prime }(x)=({\sin }^{2}x{)}^{\prime }$

$\boxed{f(x)=u^2\Rightarrow f^{\prime }(x)=2.u.u^{\prime }}$

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

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

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

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

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

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

$\boxed{\frac{\cos x}{\sin x}=\cot x}$

$\frac{f^{\prime }(x)}{f(x)}=2.\cot x$