Cos u(x)'in Türevi

Cos u(x)'in Türevi Nedir ?

Cos u(x)'in türevi, -u'(x).sin u(x)'tir.

$\frac{d}{dx}[cosu(x)]=-\frac{d}{dx}[u(x)].sinu(x)$

Cos u(x)'in Türevinin İspatı

1. Yol

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$[cosu(x){]}^{\prime }=-u^{\prime }(x).sinu(x)$

2. Yol

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

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

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

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

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

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

$[cosu(x){]}^{\prime }=0-\frac{2.u(x).u^{\prime }(x)}{2!}+\frac{4.[u(x){]}^3.u^{\prime }(x)}{4!}-\frac{6.[u(x){]}^5.u^{\prime }(x)}{6!}+\frac{8.[u(x){]}^7.u^{\prime }(x)}{8!}-...$

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

$[cosu(x){]}^{\prime }=-\frac{u(x).u^{\prime }(x)}{1}+\frac{[u(x){]}^3.u^{\prime }(x)}{3!}-\frac{[u(x){]}^5.u^{\prime }(x)}{5!}+\frac{[u(x){]}^7.u^{\prime }(x)}{7!}-...$

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

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

$[cosu(x){]}^{\prime }=-u^{\prime }(x).sinu(x)$

$Soru:$

$f(x)=cos(3x^3-2x^2)\Rightarrow f^{\prime }(x)=?$

$Cevap:$

$f(x)=cos(3x^3-2x^2)$

$f^{\prime }(x)=[cos(3x^3-2x^2){]}^{\prime }$

$\boxed{f(x)=cosu(x)\Rightarrow f^{\prime }(x)=-u^{\prime }(x).sinu(x)}$

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

$f^{\prime }(x)=-(\mathrm{3.3.}{x}^2-\mathrm{2.2.}x).sin(3x^3-2x^2)$

$f^{\prime }(x)=-(9x^2-4x).sin(3x^3-2x^2)$