derivative of square root x (derivative of root x)

what is the derivative of square root x ? (what is the derivative of root x ?)

$f(x)=\sqrt{x}\Rightarrow f^{\prime }(x)=\frac{1}{2\sqrt{x}}$

proof of derivative of square root x (proof of derivative of root x)

Way 1

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

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

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

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

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

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

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

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

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

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

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

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

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

Way 2

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

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

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

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

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

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

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

$2.f(x).f^{\prime }(x)=1$

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

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

Way 3

$\sqrt{x}=x^{\frac{1}{2}}$

$ln\sqrt{x}=ln{x}^{\frac{1}{2}}$

$ln\sqrt{x}=\frac{1}{2}.lnx$

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

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

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

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

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

$\frac{(\sqrt{x}{)}^{\prime }}{\sqrt{x}}=\frac{1.1}{2.x}$

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

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

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

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

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

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

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

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