How do I us the Limit definition of derivative on f(x)= 1/x?

Oct 3, 2014

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

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

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

Make the substitutions for $f \left(x\right)$ and $f \left(x + h\right)$

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

Find the least common denominator

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

Simplify the numerator of the complex fraction

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

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

Distribute the negative in the numerator

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

Simplify the numerator

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

Division is equivalent to multiplying by the reciprocal

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

Cancel the factors of $h$ and simplify

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

Substitute in the value of 0 for $h$ and simplify

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

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

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

See the video below.