Bölümün Türevi

İki Fonksiyonun Bölümünün Türevi Nedir ?

$[\frac{u(x)}{v(x)}{]}^{\prime }=\frac{u(x{)}^{\prime }.v(x)-v(x{)}^{\prime }.u(x)}{[v(x){]}^2}$

$\frac{d}{dx}[\frac{u(x)}{v(x)}]=\frac{\frac{d}{dx}[u(x)].v(x)-\frac{d}{dx}[v(x)].u(x)}{[v(x){]}^2}$

Bölümün Türevinin İspatı

1. Yol

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

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

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

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

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

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

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

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

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

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

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

$[\frac{u(x)}{v(x)}{]}^{\prime }=\frac{v(x)}{v(x).v(x+0)}.u^{\prime }(x)+\frac{u(x)}{v(x).v(x+0)}.-v^{\prime }(x)$

$[\frac{u(x)}{v(x)}{]}^{\prime }=\frac{v(x)}{v(x).v(x)}.u^{\prime }(x)+\frac{u(x)}{v(x).v(x)}.-v^{\prime }(x)$

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

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

$[\frac{u(x)}{v(x)}{]}^{\prime }=\frac{v(x).u^{\prime }(x)-u(x).v^{\prime }(x)}{[v(x){]}^2}$

$[\frac{u(x)}{v(x)}{]}^{\prime }=\frac{u^{\prime }(x).v(x)-v^{\prime }(x).u(x)}{[v(x){]}^2}$

2. Yol

$f(x)=\frac{u(x)}{v(x)}$

$lnf(x)=ln[\frac{u(x)}{v(x)}]$

$lnf(x)=lnu(x)-lnv(x)$

$\boxed{(lnu{)}^{\prime }=\frac{u^{\prime }}{u}}$

$\frac{f^{\prime }(x)}{f(x)}=\frac{u^{\prime }(x)}{u(x)}-\frac{v^{\prime }(x)}{v(x)}$

$\frac{f^{\prime }(x)}{f(x)}=\frac{u^{\prime }(x).v(x)-v^{\prime }(x).u(x)}{u(x).v(x)}$

$\frac{f^{\prime }(x)}{\frac{u(x)}{v(x)}}=\frac{u^{\prime }(x).v(x)-v^{\prime }(x).u(x)}{u(x).v(x)}$

$f^{\prime }(x)=\frac{\cancel{u(x)}}{v(x)}.\frac{u^{\prime }(x).v(x)-v^{\prime }(x).u(x)}{\cancel{u(x)}.v(x)}$

$f^{\prime }(x)=\frac{u^{\prime }(x).v(x)-v^{\prime }(x).u(x)}{v(x).v(x)}$

$f^{\prime }(x)=\frac{u^{\prime }(x).v(x)-v^{\prime }(x).u(x)}{[v(x){]}^2}$

3. Yol

$y=f(x)$

$y+dy=f(x+dx)$

$\cancel{y}+dy-\cancel{y}=f(x+dx)-f(x)$

$dy=f(x+dx)-f(x)$

$\frac{dy}{dx}=\frac{f(x+dx)-f(x)}{dx}$

$\frac{dy}{dx}=\underset{dx\to 0}{\lim }\frac{f(x+dx)-f(x)}{dx}$

$\frac{dy}{dx}=f^{\prime }(x)$

$\frac{dy}{dx}=y^{\prime }$

$y=\frac{u}{v}$

$y+dy=\frac{u+du}{v+dv}$

$\cancel{y}+dy-\cancel{y}=\frac{u+du}{v+dv}-\frac{u}{v}$

$dy=\frac{u+du}{v+dv}-\frac{u}{v}$

$dy=\frac{v.(u+du)-u.(v+dv)}{v.(v+dv)}$

$dy=\frac{\cancel{v.u}+v.du-\cancel{u.v}-u.dv}{v^2+v.dv}$

$dy=\frac{v.du-u.dv}{v^2+\underset{\text{ihmal edilir}}{\underbrace{v.dv}}}$

$dy=\frac{v.du-u.dv}{v^2}$

$\frac{dy}{dx}=\frac{\frac{v.du}{dx}-\frac{u.dv}{dx}}{v^2}$

$\frac{dy}{dx}=\frac{v.\frac{du}{dx}-u.\frac{dv}{dx}}{v^2}$

$y^{\prime }=\frac{v.u^{\prime }-u.v^{\prime }}{v^2}$

$y^{\prime }=\frac{u^{\prime }.v-v^{\prime }.u}{v^2}$

Soru

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

Cevap

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

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

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

$f^{\prime }(x)=\frac{3.2x-3.3-2.3x-2.5}{(2x-3{)}^2}$

$f^{\prime }(x)=\frac{\cancel{6x}-9-\cancel{6x}-10}{(2x-3{)}^2}$

$f^{\prime }(x)=\frac{-19}{(2x-3{)}^2}$

$f^{\prime }(x)=-\frac{19}{(2x-3{)}^2}$