\frac12x^{-\frac12} is the derivative of square root of x . Different methods of differentiation can be used to calculate this derivative, such as the chain rule, the power rule, and the first principle of derivatives. A derivative of square root of x can be written mathematically as \frac d{dx}\sqrt x=\frac12x^{-\frac12} or =\frac1{2\sqrt x} , using the power rule \frac d{dx}x^n=nx^{n-1} . The derivative of the square root of x can be obtained by using this formula and substituting n = {\frac12} .

**What is the Derivative of the Square Root of x?**

The derivative of square root of x is, \frac12x^{-\frac12} . We know, that in mathematics the derivative of a function is the process to find out the rate of change of a function with respect to a variable. We can find the derivative of root x by two methods

**Power rule of differentiation****By the first Principle of differentiation**

**Derivative of Square Root of x Using Power Rule**:

Power Rule Formula for derivative is:

\frac d{d_x}x^n=nx^{n-1}, .

Square Root of x is an exponential function with x as the base and n = {\frac12} as the power.

** Now,**

If substitute n = {\frac12} in the formula

\frac d{dx}x^n=nx^{n-1} ,

then

\frac d{dx}x^\frac12=\frac12x^{\frac12-1} .

\frac d{dx}x^\frac12=\frac12x^{-\frac12} .

\frac d{dx}x^\frac12 =\frac1{2\sqrt x} .

Hence, Proved the derivative of square root of x is equal to =\frac1{2\sqrt x} .

**Derivative of square Root of x Using First Principle**:

f(x)\; = \;\sqrt x

f'(x)\;=\;\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}h .

f'(x)\;=\;\lim_{h\rightarrow0}\frac{\sqrt{x+h}-\sqrt x}h .

f'(x)\;=\;\lim_{h\rightarrow0}\frac{\sqrt{x+h}-\sqrt x}h\times\frac{\sqrt{x+h}\;+\sqrt x}{\sqrt{x+h}\;+\sqrt x} .

f'(x)\;=\;\lim_{h\rightarrow0}\frac{(x+h)-x}{\sqrt{x+h}+\sqrt x} .

f'(x)\;=\;\lim_{h\rightarrow0}\frac h{h(\sqrt{x+h}+\sqrt x} .

f'(x)\;=\;\lim_{h\rightarrow0}\frac1{(\sqrt{x+h}+\sqrt x} .

f'(x)\;=\;\frac1{(\sqrt{x+0}+\sqrt x} .

f'(x)\;=\;\frac1{2\sqrt x} .