derivative of radical functions

poof of derivative of radical functions

$f(x)=\sqrt[n]{g(x)}\Rightarrow f^{\prime }(x)=\frac{g^{\prime }(x)}{n\sqrt[n]{(g(x){)}^{n-1}}}(n>1:n\in N^{+},g(x)>0)$. We can prove why the derivative of $\sqrt[n]{g(x)}$ is equal to $\frac{g^{\prime }(x)}{n\sqrt[n]{(g(x){)}^{n-1}}}$ in several ways, some of which we will show below.

Way 1

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Way 2

$f(x)=\sqrt[n]{g(x)}$

$[f(x){]}^n=(\sqrt[n]{g(x)}{)}^n$

$[f(x){]}^n=\sqrt[n]{[g(x){]}^n}$

$[f(x){]}^n=g(x)$

$\{[f(x){]}^n{\}}^{\prime }=[g(x){]}^{\prime }$

$\{[f(x){]}^n{\}}^{\prime }=g^{\prime }(x)$

$\boxed{f(x)=[g(x){]}^n\Rightarrow f^{\prime }(x)=n.[g(x){]}^{n-1}.g^{\prime }(x)}$

$n.[f(x){]}^{n-1}.f^{\prime }(x)=g^{\prime }(x)$

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

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

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

Way 3

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

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

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

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

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

$\boxed{f(x)=lng(x)\Rightarrow f^{\prime }(x)=\frac{g^{\prime }(x)}{g(x)}}$

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

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

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

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

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

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

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

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

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