# How do you use the definition of a derivative to find f' given f(x)=1/x^2 at x=1?

Jan 16, 2018

$- 2$

#### Explanation:

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

The derivative is defined as:

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

Which in our case would be:

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

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

Substitute $x = 1$:

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

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

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

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

Evaluate the limit:

$= - \frac{2}{1} = - 2$