derivative of square root functions

proof of derivative of square root functions

$f(x)=\sqrt{g(x)}\Rightarrow f^{\prime }(x)=\frac{g^{\prime }(x)}{\sqrt{g(x)}}(g(x)>0)$. We can prove why the derivative of $\sqrt{g(x)}$ is equal to $\frac{g^{\prime }(x)}{2\sqrt{g(x)}}$ in several ways, some of which we will show below.

Way 1

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Way 2

$f(x)=\sqrt{g(x)}$

$[f(x){]}^2=(\sqrt{g(x)}{)}^2$

$[f(x){]}^2=\sqrt{[g(x){]}^2}$

$[f(x){]}^2=g(x)$

$\{[f(x){]}^2{\}}^{\prime }=[g(x){]}^{\prime }$

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

$2.[f(x){]}^{2-1}.f(x{)}^{\prime }=g^{\prime }(x)$

$2.f(x).f(x{)}^{\prime }=g^{\prime }(x)$

$f^{\prime }(x)=\frac{g^{\prime }(x)}{2.f(x)}$

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

Way 3

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

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

$ln\sqrt{g(x)}=\frac{1}{2}.lng(x)$

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

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

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

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

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

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

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

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

$(\sqrt{g(x)}{)}^{\prime }=\frac{\sqrt{g(x)}.g^{\prime }(x)}{2.\sqrt{[g(x){]}^2}}$

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

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

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