# How do you find the derivative of f(x) = -2sqrt(x) by the limit process?

## f(x)=$- 2 \sqrt{x}$

$f ' \left(x\right) = - \frac{1}{\setminus} \sqrt{x}$

#### Explanation:

Given function: $f \left(x\right) = - 2 \setminus \sqrt{x}$ then its derivative using first principle as follows

$\setminus \frac{d}{\mathrm{dx}} f \left(x\right) = \setminus {\lim}_{h \setminus \to 0} \setminus \frac{f \left(x + h\right) - f \left(x\right)}{h}$

$f ' \left(x\right) = \setminus {\lim}_{h \setminus \to 0} \setminus \frac{- 2 \setminus \sqrt{x + h} - \left(- 2 \setminus \sqrt{x}\right)}{h}$

$= \setminus {\lim}_{h \setminus \to 0} \setminus \frac{- 2 \setminus \sqrt{x} \left(\setminus \sqrt{\setminus \frac{x + h}{x}} - 1\right)}{h}$

$= - 2 \setminus \sqrt{x} \setminus {\lim}_{h \setminus \to 0} \setminus \frac{\setminus \sqrt{\left(1 + \frac{h}{x}\right)} - 1}{h}$

$= - 2 \setminus \sqrt{x} \setminus {\lim}_{h \setminus \to 0} \setminus \frac{{\left(1 + \frac{h}{x}\right)}^{\frac{1}{2}} - 1}{h}$

$= - 2 \setminus \sqrt{x} \setminus {\lim}_{h \setminus \to 0} \setminus \frac{\left(1 + \left(\frac{1}{2}\right) \frac{h}{x} + \left(\frac{1}{2}\right) \left(\frac{1}{2} - 1\right) {\left(\frac{h}{x}\right)}^{2} + \setminus \ldots\right) - 1}{h}$

$= - 2 \setminus \sqrt{x} \setminus {\lim}_{h \setminus \to 0} \setminus \frac{\left(\frac{1}{2}\right) \frac{h}{x} + \left(\frac{1}{2}\right) \left(- \frac{1}{2}\right) {\left(\frac{h}{x}\right)}^{2} + \setminus \ldots}{h}$

$= - 2 \setminus \sqrt{x} \setminus {\lim}_{h \setminus \to 0} \left(\left(\frac{1}{2}\right) \frac{1}{x} + \left(\frac{1}{2}\right) \left(- \frac{1}{2}\right) \frac{h}{x} ^ 2 + \setminus \ldots\right)$

$= - 2 \setminus \sqrt{x} \left(\left(\frac{1}{2}\right) \frac{1}{x} + 0\right)$

$= - \frac{1}{\setminus} \sqrt{x}$

Jul 7, 2018

By definition:

${\left(- 2 \sqrt{x}\right)}^{'} = {\lim}_{h \to 0} \frac{- 2 \sqrt{x + h} - \left(- 2\right) \sqrt{x}}{h}$

Let: $X = x + h$

$= - 2 {\lim}_{X \to x} \frac{\sqrt{X} - \sqrt{x}}{X - x} \cdot \frac{\sqrt{X} + \sqrt{x}}{\sqrt{X} + \sqrt{x}}$

$= - 2 {\lim}_{X \to x} \frac{X - x}{X - x} \cdot \frac{1}{\sqrt{X} + \sqrt{x}}$

$= - 2 {\lim}_{h \to 0} \frac{1}{\sqrt{x + h} + \sqrt{x}} = - \frac{1}{\sqrt{x}}$