Note that the square root of 1+x can be written as (1+x)^{1/2}. In this post, we will find the derivative of the square root of 1+x by the first principle of derivatives. Using this principle, the derivative of a function f(x) is

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

## Derivative of sqrt(1+x) using Limit Definition

Let $f(x)=\sqrt{1+x}$. Then by the first principle of derivatives, the derivative of the square root of $1+x$ is given by

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

Rationalizing the numerator of the fraction involved in the above limit, the above is

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

$=\lim\limits_{h \to 0}$ $\dfrac{(\sqrt{1+x+h})^2-(\sqrt{1+x})^2}{h(\sqrt{1+x+h}+\sqrt{1+x})}$, obtained by applying the formula of $(a-b)(a+b)=a^2-b^2$

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

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

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

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

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

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

Thus the derivative of the square root of 1+x is equal to $\dfrac{1}{2\sqrt{1+x}}$ and this is obtained by the first principle of derivatives.

**Also Read: **

**Derivative of log(sin x) from first principle**

**Derivative of log(cos x) from first principle**

**Derivative of root sin x from first principle**

**Derivative of root cos x from first principle**

## FAQs

**Q1: What is the derivative of square root of 1+x?**

**Answer:** The derivative of the square root of 1+x is 1/(2 sqrt{1+x}).