Köklü İfadelerin Türevi

Köklü İfadelerin Türevi Nedir ?

ⁿ√u'nun türevi u'/(n.ⁿ√uⁿ⁻¹)'dir.

$(\sqrt[n]{u}{)}^{\prime }=\frac{u^{\prime }}{n.\sqrt[n]{u^{n-1}}}$

$\frac{d}{dx}(\sqrt[n]{u})=\frac{u^{\prime }}{n.\sqrt[n]{u^{n-1}}}$

Köklü İfadelerin Türevinin İspatı

1. Yol

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

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

$(\sqrt[n]{u(x)}{)}^{\prime }=\underset{h\to 0}{\lim }\frac{(\sqrt[n]{u(x+h)}-\sqrt[n]{u(x)}).(\sqrt[n]{[u(x+h){]}^{n-1}}+\sqrt[n]{[u(x+h){]}^{n-2}}.\sqrt[n]{u(x)}+\sqrt[n]{[u(x+h){]}^{n-3}}.\sqrt[n]{[u(x){]}^2}+...)}{h.(\sqrt[n]{[u(x+h){]}^{n-1}}+\sqrt[n]{[u(x+h){]}^{n-2}}.\sqrt[n]{u(x)}+\sqrt[n]{[u(x+h){]}^{n-3}}.\sqrt[n]{[u(x){]}^2}+...)}$

$\boxed{(a-b).(a^{n-1}+a^{n-2}.b+a^{n-3}.b^2+...+a^2.b^{n-3}+a.b^{n-2}+b^{n-1})=a^n-b^n}$

$(\sqrt[n]{u(x)}{)}^{\prime }=\underset{h\to 0}{\lim }\frac{\sqrt[n]{[u(x+h){]}^n}-\sqrt[n]{[u(x){]}^n}}{h.(\sqrt[n]{[u(x+h){]}^{n-1}}+\sqrt[n]{[u(x+h){]}^{n-2}}.\sqrt[n]{u(x)}+\sqrt[n]{[u(x+h){]}^{n-3}}.\sqrt[n]{[u(x){]}^2}+...)}$

$(\sqrt[n]{u(x)}{)}^{\prime }=\underset{h\to 0}{\lim }\frac{u(x+h)-u(x)}{h.(\sqrt[n]{[u(x+h){]}^{n-1}}+\sqrt[n]{[u(x+h){]}^{n-2}}.\sqrt[n]{u(x)}+\sqrt[n]{[u(x+h){]}^{n-3}}.\sqrt[n]{[u(x){]}^2}+...)}$

$(\sqrt[n]{u(x)}{)}^{\prime }=\underset{h\to 0}{\lim }[\frac{u(x+h)-u(x)}{h}.\frac{1}{\sqrt[n]{[u(x+h){]}^{n-1}}+\sqrt[n]{[u(x+h){]}^{n-2}}.\sqrt[n]{u(x)}+\sqrt[n]{[u(x+h){]}^{n-3}}.\sqrt[n]{[u(x){]}^2}+...}]$

$(\sqrt[n]{u(x)}{)}^{\prime }=\underset{h\to 0}{\lim }\frac{u(x+h)-u(x)}{h}.\underset{h\to 0}{\lim }\frac{1}{\sqrt[n]{[u(x+h){]}^{n-1}}+\sqrt[n]{[u(x+h){]}^{n-2}}.\sqrt[n]{u(x)}+\sqrt[n]{[u(x+h){]}^{n-3}}.\sqrt[n]{[u(x){]}^2}+...}$

$(\sqrt[n]{u(x)}{)}^{\prime }=u^{\prime }(x).\frac{1}{\sqrt[n]{[u(x+0){]}^{n-1}}+\sqrt[n]{[u(x+0){]}^{n-2}}.\sqrt[n]{u(x)}+\sqrt[n]{[u(x+0){]}^{n-3}}.\sqrt[n]{[u(x){]}^2}+...}$

$(\sqrt[n]{u(x)}{)}^{\prime }=u^{\prime }(x).\frac{1}{\sqrt[n]{[u(x){]}^{n-1}}+\sqrt[n]{[u(x){]}^{n-2}}.\sqrt[n]{u(x)}+\sqrt[n]{[u(x){]}^{n-3}}.\sqrt[n]{[u(x){]}^2}+...}$

$(\sqrt[n]{u(x)}{)}^{\prime }=u^{\prime }(x).\frac{1}{\sqrt[n]{[u(x){]}^{n-1}}+\sqrt[n]{[u(x){]}^{n-2}.u(x)}+\sqrt[n]{[u(x){]}^{n-3}.[u(x){]}^2}+...}$

$(\sqrt[n]{u(x)}{)}^{\prime }=u^{\prime }(x).\frac{1}{\sqrt[n]{[u(x){]}^{n-1}}+\sqrt[n]{[u(x){]}^{n-2+1}}+\sqrt[n]{[u(x){]}^{n-3+2}}+...}$

$(\sqrt[n]{u(x)}{)}^{\prime }=u^{\prime }(x).\frac{1}{\underset{\text{n tane}}{\underbrace{\sqrt[n]{[u(x){]}^{n-1}}+\sqrt[n]{[u(x){]}^{n-1}}+\sqrt[n]{[u(x){]}^{n-1}}+...}}}$

$(\sqrt[n]{u(x)}{)}^{\prime }=u^{\prime }(x).\frac{1}{n.\sqrt[n]{[u(x){]}^{n-1}}}$

$(\sqrt[n]{u(x)}{)}^{\prime }=\frac{u^{\prime }(x)}{n.\sqrt[n]{[u(x){]}^{n-1}}}$

$u(x)=u$

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

$(\sqrt[n]{u}{)}^{\prime }=\frac{u^{\prime }}{n.\sqrt[n]{u^{n-1}}}$

2. Yol

$y=\sqrt[n]{u(x)}$

$y^n=(\sqrt[n]{u(x)}{)}^n$

$y^n=u(x)$

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

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

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

$y^{\prime }=\frac{u^{\prime }(x)}{n.y^{n-1}}$

$y^{\prime }=\frac{u^{\prime }(x)}{n.(\sqrt[n]{u(x)}{)}^{n-1}}$

$y^{\prime }=\frac{u^{\prime }(x)}{n.\sqrt[n]{[u(x){]}^{n-1}}}$

$(\sqrt[n]{u(x)}{)}^{\prime }=\frac{u^{\prime }(x)}{n.\sqrt[n]{[u(x){]}^{n-1}}}$

$u(x)=u$

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

$(\sqrt[n]{u}{)}^{\prime }=\frac{u^{\prime }}{n.\sqrt[n]{u^{n-1}}}$

3. Yol

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

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

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

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

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

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

$\frac{(\sqrt[n]{u(x)}{)}^{\prime }}{\sqrt[n]{u(x)}}=\frac{u^{\prime }(x)}{n.u(x)}$

$\frac{(\sqrt[n]{u(x)}{)}^{\prime }}{\frac{\cancel{\sqrt[n]{u(x)}}}{\cancel{\sqrt[n]{u(x)}}}}=\frac{1}{\frac{n.\cancel{u(x)}\sqrt[n]{[u(x){]}^{n-1}}}{\cancel{\sqrt[n]{u(x)}}}}$

$(\sqrt[n]{u(x)}{)}^{\prime }=\frac{u^{\prime }(x)}{n.\sqrt[n]{[u(x){]}^{n-1}}}$

$u(x)=u$

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

$(\sqrt[n]{u}{)}^{\prime }=\frac{u^{\prime }}{n.\sqrt[n]{u^{n-1}}}$