Derivative of root(x)-1

Q: Q1: What is the derivative of √x -1?

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

The derivative of root x minus 1 is 1 by 2 root x. In this post, we will prove this derivative formula by the power rule of derivatives as well as by the first principle of derivatives.

Table of Contents

Derivative of root(x)-1 Formula

The Formula for the derivative of root x -1 is given below:

$\dfrac{d}{dx}(\sqrt{x}-1)=\dfrac{1}{2\sqrt{x}}$ or
$(\sqrt{x}-1)^\prime=\dfrac{1}{2\sqrt{x}}$.

Derivative of root(x)-1 by Power Rule

The power rule of derivatives: If n is the exponent of x, then the derivative of $x^n$ is

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

Note that $\sqrt{x}$ can be written as x raised to $\frac{1}{2}$. So

$\dfrac{d}{dx}(\sqrt{x}-1)$ $=\dfrac{d}{dx}(\sqrt{x})-\dfrac{d}{dx}(1)$

$=\dfrac{d}{dx}(x^{\frac{1}{2}})-0$ as the derivative of a constant is zero.

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

Applying the above power rule with exponent $n=\dfrac{1}{2}$, the above is

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

Simplifying, it is

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

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

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

So the derivative of sqrt(x)-1 is 1/2root(x) and this is obtained by the power rule of derivatives.

Derivative of root(x)-1 by First Principle

The derivative of a function f(x) from first principle is $\dfrac{d}{dx}(f(x))$ $=\lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$. We take $f(x)=\sqrt{x}-1$. So the derivative of root(x)-1 by first principle is

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

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

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

$=\lim\limits_{h \to 0} [\dfrac{(\sqrt{x+h})^2-(\sqrt{x})^2}{h}$ $\times \dfrac{1}{\sqrt{x+h}+\sqrt{x}}]$ as we know that $(x-y)(x+y)$ $=x^2-y^2$.

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

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

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

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

This shows that the derivative of $\sqrt{x}-1$ is equal to $\dfrac{1}{2\sqrt{x}}$ which is obtained by the first principle of derivatives.

Also Read:

Derivative of root x

Derivative of x sin x