[katex] \frac12x^{-\frac12} [/katex] is the derivative of the square root of [katex] x [/katex]. 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 [katex] \frac d{dx}\sqrt x=\frac12x^{-\frac12} [/katex] or [katex] =\frac1{2\sqrt x} [/katex], using the power rule [katex] \frac d{dx}x^n=nx^{n-1} [/katex]. The derivative of the square root of x can be obtained by using this formula and substituting [katex] n = {\frac12} [/katex].
What is the Derivative of the Square Root of x?
The derivative of square root of x is, [katex] \frac12x^{-\frac12} [/katex] . 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 the derivative is:
[katex] \frac d{d_x}x^n=nx^{n-1}, [/katex].
Square Root of x is an exponential function with x as the base and [katex] n = {\frac12} [/katex] as the power.
Now,
If substitute [katex] n = {\frac12} [/katex] in the formula
[katex] \frac d{dx}x^n=nx^{n-1} [/katex],
then
[katex] \frac d{dx}x^\frac12=\frac12x^{\frac12-1} [/katex].
[katex] \frac d{dx}x^\frac12=\frac12x^{-\frac12} [/katex].
[katex] \frac d{dx}x^\frac12 =\frac1{2\sqrt x} [/katex].
Hence, Proved the derivative of square root of x is equal to [katex] =\frac1{2\sqrt x} [/katex].
Derivative of square Root of x Using First Principle:
[katex] f(x)\; = \;\sqrt x [/katex]
[katex] f'(x)\;=\;\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}h [/katex].
[katex] f'(x)\;=\;\lim_{h\rightarrow0}\frac{\sqrt{x+h}-\sqrt x}h [/katex].
[katex] f'(x)\;=\;\lim_{h\rightarrow0}\frac{\sqrt{x+h}-\sqrt x}h\times\frac{\sqrt{x+h}\;+\sqrt x}{\sqrt{x+h}\;+\sqrt x} [/katex].
[katex] f'(x)\;=\;\lim_{h\rightarrow0}\frac{(x+h)-x}{\sqrt{x+h}+\sqrt x} [/katex].
[katex] f'(x)\;=\;\lim_{h\rightarrow0}\frac h{h(\sqrt{x+h}+\sqrt x} [/katex].
[katex] f'(x)\;=\;\lim_{h\rightarrow0}\frac1{(\sqrt{x+h}+\sqrt x} [/katex].
[katex] f'(x)\;=\;\frac1{(\sqrt{x+0}+\sqrt x} [/katex].
[katex] f'(x)\;=\;\frac1{2\sqrt x} [/katex].