Derivative of Root x by First Principle

Derivative of root x: The square root of x is a very important function in Mathematics. In this post, we will find the derivative of the square root of x using the first principle of derivatives and by the power rule of derivatives.

At first, we find the derivative of root x by limit definition, that is, calculate the derivative of y=√x using first principle.

Derivative of root x by First Principle

The first principle of derivatives says that the derivative of a function $f(x)$ is given by

$\dfrac{d}{dx}(f(x))$ $=\lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$

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

So we get the derivative of the square root of $x$ is

$\dfrac{d}{dx}(\sqrt{x})$ $=\lim\limits_{h \to 0} \dfrac{\sqrt{x+h}-\sqrt{x}}{h}$

Now we will rationalize the numerator of the \dfraction involved in the above limit. So we get

$=\lim\limits_{h \to 0}$ $[\dfrac{\sqrt{x+h}-\sqrt{x}}{h}$ $\times \dfrac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}]$

Applying the formula of $(a-b)(a+b)=a^2-b^2$ we get that

$=\lim\limits_{h \to 0}$ $\dfrac{(\sqrt{x+h})^2-(\sqrt{x})^2}{h(\sqrt{x+h}+\sqrt{x})}$

$=\lim\limits_{h \to 0}$ $\dfrac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})}$

$=\lim\limits_{h \to 0}$ $\dfrac{h}{h(\sqrt{x+h}+\sqrt{x})}$

Canceling $h$ from the numerator and the denominator, we get that

$=\lim\limits_{h \to 0}$ $\dfrac{1}{\sqrt{x+h}+\sqrt{x}}$

$=\dfrac{1}{\sqrt{x+0}+\sqrt{x}}$

$=\dfrac{1}{\sqrt{x}+\sqrt{x}}$

$=\dfrac{1}{2\sqrt{x}}$

So the derivative of root x is 1/2root(x), and this is obtained by the first principle of derivatives.

Next, we will evaluate the derivative of root x by the power rule of derivatives. The rule is given below:

$\dfrac{d}{dx}(x^n)=nx^{n-1}$

Derivative of root x by Power Rule

By the rule of indices, we can write $\sqrt{x}$ as $x^{1/2}$. So we have

$\dfrac{d}{dx}(\sqrt{x})$ $=d/dx(x^{1/2})$

$=\dfrac{1}{2} x^{1/2-1}$ by the above power rule of derivatives.

$=\dfrac{1}{2} x^{-1/2}$

$=\dfrac{1}{2x^{1/2}}$

$=\dfrac{1}{2\sqrt{x}}$

So the derivative of $\sqrt{x}$ is $\dfrac{1}{2\sqrt{x}}$, and this is obtained by the power rule of derivatives.