# How do I use the definition of a derivative to find the derivative of f(x)=1/x at x=3?

Oct 1, 2014

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

$f \left(x\right) = \frac{1}{x}$
$f \left(x + h\right) = \frac{1}{x + h}$

Substitute in these values

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

Get common denominator for the numerator of the complex fraction.

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{\frac{x}{x} \cdot \frac{1}{x + h} - \frac{1}{x} \cdot \frac{x + h}{x + h}}{h}$

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

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

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

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{- h}{x \left(x + h\right)} \cdot \frac{1}{h}$

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

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

$f ' \left(x\right) = \frac{- 1}{x \left(x + 0\right)}$

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

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

$f ' \left(3\right) = \frac{- 1}{{\left(3\right)}^{2}} = - \frac{1}{9}$

Alternative method

Now take the derivative of $f \left(x\right)$ using the power rule.

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

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

$f ' \left(3\right) = - \frac{1}{3} ^ 2 = - \frac{1}{9}$