Çarpımın Türevi

İki Fonksiyonun Çarpımının Türevi Nedir ?

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

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

Çarpımın Türevinin İspatı

1. Yol

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

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

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

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

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

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

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

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

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

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

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

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

2. Yol

$f(x)=u(x).v(x)$

$lnf(x)=ln[u(x).v(x)]$

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

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

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

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

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

3. Yol

$y=f(x)$

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

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

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

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

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

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

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

$y=u.v$

$y+dy=(u+du).(v+dv)$

$y+dy=u.v+u.dv+v.du+du.dv$

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

$dy=u.dv+v.du+du.dv$

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

$dy=u.dv+v.du$

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

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

$y^{\prime }=u.v^{\prime }+v.u^{\prime }$

$y^{\prime }=u^{\prime }.v+v^{\prime }.u$

4. Yol

$(u+v{)}^2=u^2+2.u.v+v^2$

$(u+v{)}^2-u^2-v^2=2.u.v$

$\frac{(u+v{)}^2-u^2-v^2}{2}=\frac{\cancel{2}.u.v}{\cancel{2}}$

$u.v=\frac{(u+v{)}^2-u^2-v^2}{2}$

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

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

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

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

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

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

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

$(u.v{)}^{\prime }=\frac{2.u.v^{\prime }+2.v.u^{\prime }}{2}$

$(u.v{)}^{\prime }=\frac{\cancel{2}.(u.v^{\prime }+v.u^{\prime })}{\cancel{2}}$

$(u.v{)}^{\prime }=u.v^{\prime }+v.u^{\prime }$

$(u.v{)}^{\prime }=u^{\prime }.v+v^{\prime }.u$

Soru

$f(x)=(3x-4).(5x+7)\Rightarrow f^{\prime }(x)=?$

Cevap

$f^{\prime }(x)=[(3x-4).(5x+7){]}^{\prime }$

$f^{\prime }(x)=(3x-4{)}^{\prime }.(5x+7)+(5x+7{)}^{\prime }.(3x-4)$

$f^{\prime }(x)=3.(5x+7)+5.(3x-4)$

$f^{\prime }(x)=3.5x+3.7+5.3x-5.4$

$f^{\prime }(x)=15x+21+15x-20$

$f^{\prime }(x)=15x+15x+21-20$

$f^{\prime }(x)=30x+1$