Question

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

Answer 1

`f'(x) = (-2x)/((x^2-1)(x^2-1)) = (-2x)/(x^2-1)^2`
Please refer to explanation below for more information.

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)

`(dy)/(dx) = f'(x) = lim_(h->0) (f(x+h)-f(x))/h`

Here is how:

Step 1: Let's set it up, with the given function `f(x)= 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'(x) = lim_(h->0)(color(red)((1/((x+h)^2-1)))- (1/(x^2-1)))/h`

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

Find the least common denominator

`f'(x) = lim_(h->0)[( 1(x^2-1)-((x+h)^2-1))/((x^2-1)((x+h)^2-1))]/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->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->0) (-2x-0)/((x^2-1)[(x+0)^2-1])`

`f'(x) = (-2x)/((x^2-1)(x^2-1)) = (-2x)/(x^2-1)^2`