Question

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

Answer 1

Long divide then use limit definition of derivative to find:

`f'(x) = 1/2 - 5/(2(2x-3)^2)`

Explanation:

Long divide:

...to find...

`f(x) = (x^2-1)/(2x-3)=1/2x+3/4+5/(8x-12)`

Then using the limit definition of derivative:

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

`=lim_(h->0) ((1/2(x+h)+3/4+5/(8(x+h)-12)) - (1/2x+3/4+5/(8x-12)))/h`

`=lim_(h->0) (1/2h+(5/(8(x+h)-12)-5/(8x-12)))/h`

`=1/2+lim_(h->0) ((5/(8(x+h)-12)-5/(8x-12))/h)`

`=1/2+lim_(h->0) ((5((8x-12)-(8(x+h)-12)))/(h(8(x+h)-12)(8x-12)))`

`=1/2+lim_(h->0) ((-40h)/(h(8(x+h)-12)(8x-12)))`

`=1/2+lim_(h->0) ((-40)/((8(x+h)-12)(8x-12)))`

`=1/2-40/((8x-12)^2)`

`=1/2-40/(16(2x-3)^2)`

`=1/2-5/(2(2x-3)^2)`