Kareköklü İfadelerin Türevi

Kareköklü İfadelerin Türevi Nedir ?

√u'nun türevi u'/(2.√u)'dur.

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

$\frac{d}{dx}(\sqrt{u})=\frac{u^{\prime }}{2.\sqrt{u}}$

Kareköklü İfadelerin Türevinin İspatı

1. Yol

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

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

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

$\boxed{(a+b).(a-b)=a^2-b^2}$

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

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

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

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

$(\sqrt{u(x)}{)}^{\prime }=u^{\prime }(x).\frac{1}{\sqrt{u(x+0)}+\sqrt{u(x)}}$

$(\sqrt{u(x)}{)}^{\prime }=u^{\prime }(x).\frac{1}{\sqrt{u(x)}+\sqrt{u(x)}}$

$(\sqrt{u(x)}{)}^{\prime }=u^{\prime }(x).\frac{1}{2.\sqrt{u(x)}}$

$(\sqrt{u(x)}{)}^{\prime }=\frac{u^{\prime }(x)}{2.\sqrt{u(x)}}$

$u(x)=u$

$u^{\prime }(x)=u^{\prime }$

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

2. Yol

$y=\sqrt{u(x)}$

$y^2=(\sqrt{u(x)}{)}^2$

$y^2=u(x)$

$(y^2{)}^{\prime }=u^{\prime }(x)$

$\boxed{(u^n{)}^{\prime }=n.(u{)}^{n-1}.u^{\prime }}$

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

$y^{\prime }=\frac{u^{\prime }(x)}{2.y}$

$y^{\prime }=\frac{u^{\prime }(x)}{2.\sqrt{u(x)}}$

$(\sqrt{u(x)}{)}^{\prime }=\frac{u^{\prime }(x)}{2.\sqrt{u(x)}}$

$u(x)=u$

$u^{\prime }(x)=u^{\prime }$

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

3. Yol

$\sqrt{u(x)}=[u(x){]}^{\frac{1}{2}}$

$ln\sqrt{u(x)}=ln[u(x){]}^{\frac{1}{2}}$

$ln\sqrt{u(x)}=\frac{1}{2}.lnu(x)$

$(ln\sqrt{u(x)}{)}^{\prime }=[\frac{1}{2}.lnu(x){]}^{\prime }$

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

$\frac{(\sqrt{u(x)}{)}^{\prime }}{\sqrt{u(x)}}=\frac{1}{2}.\frac{u^{\prime }(x)}{u(x)}$

$\frac{(\sqrt{u(x)}{)}^{\prime }}{\sqrt{u(x)}}=\frac{u^{\prime }(x)}{2.u(x)}$

$\frac{(\sqrt{u(x)}{)}^{\prime }}{\frac{\cancel{\sqrt{u(x)}}}{\cancel{\sqrt{u(x)}}}}=\frac{u^{\prime }(x)}{\frac{2.\cancel{u(x)}\sqrt{u(x)}}{\cancel{\sqrt{u(x)}}}}$

$(\sqrt{u(x)}{)}^{\prime }=\frac{u^{\prime }(x)}{2.\sqrt{u(x)}}$

$u(x)=u$

$u^{\prime }(x)=u^{\prime }$

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