# How do you find the formula for the derivative of 1/x?

Apr 14, 2015

$\frac{d {x}^{a}}{\mathrm{dx}} = a \cdot {x}^{a - 1}$

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

Therefore
$\frac{d \frac{1}{x}}{\mathrm{dx}} = - {x}^{- 2}$ or $- \frac{1}{{x}^{2}}$

Apr 14, 2015

I will assume that you are working from first principles, that is the definition.

There are two choices for how to "officially" define the derivative, every textbook author and teacher makes a decision which one to make official.

I'll use: $f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$

For $f \left(x\right) = \frac{1}{x}$, we get:

$f ' \left(x\right) = {\lim}_{h \rightarrow 0} \frac{f \left(x + h\right) - f \left(x\right)}{h}$

$\textcolor{w h i t e}{\text{ssssss}}$$= {\lim}_{h \rightarrow 0} \frac{\frac{1}{x + h} - \frac{1}{x}}{h}$

$\textcolor{w h i t e}{\text{ssssss}}$$= {\lim}_{h \rightarrow 0} \frac{\frac{x - \left(x + h\right)}{x \left(x + h\right)}}{\frac{h}{1}}$

$\textcolor{w h i t e}{\text{ssssss}}$$= {\lim}_{h \rightarrow 0} \frac{\left(x - x - h\right)}{x \left(x + h\right)} \cdot \left(\frac{1}{h}\right)$

$\textcolor{w h i t e}{\text{ssssss}}$$= {\lim}_{h \rightarrow 0} \frac{\left(- 1\right)}{x \left(x + h\right)}$

$\textcolor{w h i t e}{\text{ssssss}}$$= - \frac{1}{x} ^ 2$

That is: $f ' \left(x\right) = - \frac{1}{x} ^ 2$