Sin u(x)'in Türevi

Sin u(x)'in Türevi Nedir ?

Sin u(x)'in türevi, u'(x).cos u(x)'tir.

$\frac{d}{dx}[\sin u(x)]=\frac{d}{dx}[u(x)].\cos u(x)$

Sin u(x)'in Türevinin İspatı

1. Yol

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

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

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

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

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

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

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

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

$h=\frac{u(x+h)-u(x)}{2}$$(h\to 0)$

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

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

$[\sin u(x){]}^{\prime }=1.u^{\prime }(x).\cos [\frac{u(x+0)+u(x)}{2}]$

$[\sin u(x){]}^{\prime }=1.u^{\prime }(x).\cos [\frac{u(x)+u(x)}{2}]$

$[\sin u(x){]}^{\prime }=1.u^{\prime }(x).\cos [\frac{\cancel{2}.u(x)}{\cancel{2}}]$

$[\sin u(x){]}^{\prime }=u^{\prime }(x).\cos u(x)$

2. Yol

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

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

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

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

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

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

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

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

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

$[\sin u(x){]}^{\prime }=u^{\prime }(x).\cos u(x)$

$Soru:$

$f(x)=\sin (3x^2+5x+4)\Rightarrow f^{\prime }(x)=?$

$Cevap:$

$f(x)=\sin (3x^2+5x+4)$

$f^{\prime }(x)=[\sin (3x^2+5x+4){]}^{\prime }$

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

$f^{\prime }(x)=(3x^2+5x+4{)}^{\prime }.\cos (3x^2+5x+4)$

$f^{\prime }(x)=(\mathrm{2.3.}x+5+0).\cos (3x^2+5x+4)$

$f^{\prime }(x)=(6x+5).\cos (3x^2+5x+4)$