Derivative of 1 by root x | Integration of 1/root(x)

The derivative of 1 by root x is -1/(2x root(x)). In this post, we will find the derivative of 1/root(x) by the first principle of derivatives. We will also find the integration of 1 by root x.

Derivative of 1 by root x from First Principle

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

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

Step 1: In the above formula, we put $f(x)=\dfrac{1}{\sqrt{x}}.$ Thus the derivative of the reciprocal of the square root of $x$ by the first principle is

$\frac{d}{dx}(1/\sqrt{x})$ $=\lim\limits_{h \to 0} \dfrac{1/\sqrt{x+h}-1/\sqrt{x}}{h}$

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

Step 2: We will now rationalize the numerator.

$=\lim\limits_{h \to 0}$ $\dfrac{1}{h}[\dfrac{\sqrt{x}-\sqrt{x+h}}{\sqrt{x}\sqrt{x+h}}$ $\times \dfrac{\sqrt{x}+\sqrt{x+h}}{\sqrt{x}+\sqrt{x+h}}]$

Step 3: Apply the formula of $(a-b)(a+b)=a^2-b^2$ in the numerator.

$=\lim\limits_{h \to 0}$ $\dfrac{1}{h}[\dfrac{(\sqrt{x})^2-(\sqrt{x+h})^2}{\sqrt{x}\sqrt{x+h}(\sqrt{x}+\sqrt{x+h})}]$

$=\lim\limits_{h \to 0}$ $\dfrac{1}{h}[\dfrac{x-x-h}{\sqrt{x}\sqrt{x+h}(\sqrt{x}+\sqrt{x+h})}]$

$=\lim\limits_{h \to 0}$ $\dfrac{1}{h}[\dfrac{-h}{\sqrt{x}\sqrt{x+h}(\sqrt{x}+\sqrt{x+h})}]$

$=\lim\limits_{h \to 0}$ $[\dfrac{-1}{\sqrt{x}\sqrt{x+h}(\sqrt{x}+\sqrt{x+h})}]$

$=\dfrac{-1}{\sqrt{x}\sqrt{x+0}(\sqrt{x}+\sqrt{x+0})}$

$=-\dfrac{1}{(\sqrt{x})^2 \cdot 2\sqrt{x}}$

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

So the derivative of 1 by root x is equal to -1/(2x root(x)) which is obtained by the first principle of derivatives.

Integration of Cube Root of x

Derivative of root x + 1 by root x

Derivative of x root(x)

Integration of 1 by root x

Note that 1/root(x) can be written as x to the power -1/2, that is,

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

Now, $\int \dfrac{1}{\sqrt{x}} dx=\int x^{-1/2} dx$

$=\dfrac{x^{-1/2+1}}{-1/2+1}$ by the power rule of integrals

$\int x^n dx =\frac{x^{n+1}}{n+1}$

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

$=2\sqrt{x}$

So the integration of 1 by root(x) is equal to 2 root(x).

Question: Find the integration of 1 by root(x) from 0 to 1, that is,

Find $\int_0^1 \dfrac{1}{\sqrt{x}} dx$

As we know from above that $\int \dfrac{1}{\sqrt{x}} dx = 2\sqrt{x}$, we obtain that

$\int_0^1 \dfrac{1}{\sqrt{x}} dx$

$=[2\sqrt{x}]_0^1$

$=2(\sqrt{1}-\sqrt{0})$

$=2(1-0)$ $=2$

So the value of the integration of 1/root(x) from 0 to 1 is equal to 2.

FAQs

Q1: What is the Derivative of 1/root x?

Answer: The derivative of 1/root x is -1/(2x root(x)).

Q2: What is the Integration of 1/root x?

Answer: The integration of 1/root x is 2 root(x).