Question

How do you find f'(3) using the limit definition given `1/(x^2-1)`?

Answer 1

When asked to differentiate using the limit definition, use the formula `f'(x) = lim_(h ->0) (f(x + h) - f(x))/h`.

Explanation:

Let the function be `f(x)`:

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

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

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

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

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

`f'(x) = lim_(h->0) (-2xh + h^2)/(h(x^4 + 2x^3h + x^2h^2 - x^2 - x^2-2xh-h^2 + 1))`

`f'(x) = lim_(h->0) (cancel(h)(-2x + h))/(cancel(h)(x^4 + 2x^3h + x^2h^2+ 2xh + h^2 + 1 - 2x^2)`

Substituting:

`f'(x) = (-2x + 0)/(x^4 + 2x^3(0) + x^2(0)^2 + 2x(0) + 0^2 + 1 - 2x^2)`

`f'(x) = -(2x)/(x^4 - 2x^2+ 1)`

Hence, the derivative of `f(x) = 1/(x^2 - 1)` is `f'(x) = -(2x)/(x^4 - 2x^2 + 1)`. All that we have left to do is plug in `x = 3` into our derivative:

`f'(3) = -(2 xx 3)/(3^4 - 2(3)^2 + 1)`

`f'(3) = -6/(81 - 18 + 1)`

`f'(3) = -6/(64)`

`f'(3) = -3/32`

In summary, when `f'(3)` is evaluated inside the derivative of `f(x) = 1/(x^2 - 1)`, a result of `-3/32` is obtained.

Hopefully this helps!