# How do you find the derivative of 1/x^2 ?

May 16, 2016

$- \frac{2}{x} ^ 3$

#### Explanation:

We will use the power rule, which states that the derivative of ${x}^{n}$ is $n {x}^{n - 1}$.

We can use the power rule once we write $\frac{1}{x} ^ 2$ as ${x}^{-} 2$.

Thus, according to the power rule, the derivative of ${x}^{-} 2$ is $- 2 {x}^{- 2 - 1} = - 2 {x}^{-} 3 = - \frac{2}{x} ^ 3$.

May 16, 2016

Use the limit definition to find:

$\frac{d}{\mathrm{dx}} \frac{1}{x} ^ 2 = - \frac{2}{x} ^ 3$

#### Explanation:

The power rule is good, but let's find it using the limit definition:

Let $f \left(x\right) = \frac{1}{x} ^ 2$

Then:

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

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

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

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

$= {\lim}_{h \to 0} \frac{- \textcolor{red}{\cancel{\textcolor{b l a c k}{h}}} \left(2 x + h\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{h}}} {\left(x + h\right)}^{2} {x}^{2}}$

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

$= \frac{- 2 x}{x} ^ 4$

$= - \frac{2}{x} ^ 3$