# 3.The Square Root

## 19.01.2021

### Contents/Index

1. Logarithmic
2. Absoulte Value
@3. The Square Root

The square root of $x$ is written as a function as $$f(x) = \sqrt{x}$$ It is the inverse of $g(x) = x^2$, that is $g \circ f = f \circ g = id$. Or we can state this as $$\sqrt{x^2} = \sqrt{x}^2 = x$$ The square root is defined in the range $[0,\infty]$, thus not for $x \lt 0$. We have for $x \lt 0$, that $x^2 \gt 0$. So though $f \circ g$ is defined on $[-\infty,\infty]$, we have that this compound function is not injective. It can be surjective, if defined properly: $$f \circ g : \mathbb{R} \rightarrow \mathbb{R}^{+}$$

We can plot the function with the following Python3 code:

import numpy as np import matplotlib.pyplot as plt def f(x): return np.sqrt(x) x = [float(x) / 10.0 for x in range(0,10 ** 4)] y = [f(x) for x in x] plt.plot(x,y) plt.show()

For which we get the plot as seen in Figure 1.

## The derivative of the absolute value

The derivative is given as $$f(x)' = \frac{1}{2 \sqrt{x}}$$ We can show this as follows. Rewrite $$\sqrt{x} = x^{1/2}$$ Use the power rule to obtain $$f(x)' = \frac{1}{2} x^{-1/2}$$ We have that $$x^{-1/2} = \frac{1}{x^{1/2}} = \frac{1}{\sqrt{x}}$$ So in total we have that $$f(x)' = \frac{1}{2} \cdot \frac{1}{\sqrt{x}} = \frac{1}{2 \sqrt{x}}$$