Question

How do you use the definition of a derivative to find the derivative of `f(x) = (x^2-1) / (2x-3)`?

Answer 1

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

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

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

`f'(x) = lim_(h->0) ((2x^3 + 4x^2h + 2h^2x - 2x - 3x^2 - 6xh -3h^2 + 3 - (2x^3 + 2x^2h - 3x^2 - 2x- 2h + 3))/((2x + 2h - 3)(2x -3)))/h`

`f'(x) = lim_(h->0) ((2x^3 + 4x^2h + 2h^2x - 2x - 3x^2 - 6xh - 3h^2 + 3 - 2x^3 - 2x^2h + 3x^2 + 2x + 2h - 3)/((2x + 2h - 3)(2x - 3)))/h`

`f'(x) = lim_(h->0) ((4x^2h + 2h^2x- 3h^2 - 6xh - 2x^2h+ 2h)/((2x + 2h - 3)(2x -3)))/h`

`f'(x) = lim_(h->0) ((2h^2x + 2x^2h - 3h^2 - 6xh + 2h)/((2x + 2h - 3)(2x - 3)))/h`

`f'(x) = lim_(h->0) (2h^2x + 2x^2h- 3h^2 - 6xh + 2h)/(h(2x + 2h - 3)(2x - 3))`

`f'(x)= lim_(h->0) (h(2hx + 2x^2 - 3h - 6x + 2))/(h(2x+ 2h -3)(2x - 3))`

`f'(x) = lim_(h->0) (2hx +2x^2- 3h - 6x +2)/((2x + 2h - 3)(2x -3)`

`f'(x) = (2(0) x + 2x^2 - 3(0) - 6x + 2)/((2x + 2(0) - 3)(2x - 3))`

`f'(x) = (2x^2 - 6x + 2)/(2x - 3)^2`

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

Verification using the quotient rule yields the same result.

Hopefully this helps!