# Using the limit definition, how do you find the derivative of 1/(x^2-1)?

Feb 5, 2016

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

#### Explanation:

If you apply limit to a difference quotient formula, you will get the derivative of the function using the limit of definition

Remember: The difference quotient formula is

(f(x+h)-f(x))/h ; h!= 0  (one of many form of the formula)

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

Here is how:

Step 1: Let's set it up, with the given function $f \left(x\right) = \frac{1}{{x}^{2} - 1}$

Step 2: We know, f(x+h) =color(red)( 1/((x+h)^2 -1)

Step 3: Let's set up the limit, to find the derivative

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

Step 4: Let's simplify the expression first before we evaluate the limit (here comes the ALGEBRA!!)

Find the least common denominator

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

Multiply the numerator and distribute negative one and divide the fraction to get

lim_(h->0)((x^2-1)-(x^2 +2xh+h^2-1))/((h)(x^2-1)[(x+h)^2-1]

lim_(h->0)(x^2-1-x^2 -2xh-h^2+1)/((h)(x^2-1)[(x+h)^2-1]

Simply, by combine all the like terms, and factor out the common factor on the numerator to get

lim_(h->0)(-2xh-h^2)/((h)(x^2-1)[(x+h)^2-1]

lim_(h->0)(h(2x-h))/((h)(x^2-1)[(x+h)^2-1]

Remember, $h \to 0$ doesn't not meant that $h = 0$ therefore we can divide/cross divide the numerator and denominator to get

color(blue)(lim_(h->0)(-2x-h)/((x^2-1)[(x+h)^2-1]

Then, we can directly substitute $h = 0$ , to evaluate the limit

lim_(h->0)(-2x-h)/((x^2-1)[(x+h)^2-1]

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

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