# Derivative of root(x)-1

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.

## 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.

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## FAQs

Q1: What is the derivative of √x -1?

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