Given the function `f(x)=2x^2-3x` , c=2.

We have to use the alternative form of derivative to find the derivative at x=c.

Here given that c=2.

So by definition of alternative derivative we have,

`f'(c)=lim_(x->c)(f(x)-f(c))/(x-c)`

`f'(2)=lim_(x->2)(f(x)-f(2))/(x-2)`

` =lim_(x->2)(2x^2-3x-(2(2^2)-3(2)))/(x-2)`

`=lim_(x->2)(2x^2-3x-2)/(x-2)`

`=lim_(x->2)((2x+1)(x-2))/(x-2)`

`=lim_(x->2) 2x+1`

`=5`

## Unlock

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

Given the function `f(x)=2x^2-3x` , c=2.

We have to use the alternative form of derivative to find the derivative at x=c.

Here given that c=2.

So by definition of alternative derivative we have,

`f'(c)=lim_(x->c)(f(x)-f(c))/(x-c)`

`f'(2)=lim_(x->2)(f(x)-f(2))/(x-2)`

` =lim_(x->2)(2x^2-3x-(2(2^2)-3(2)))/(x-2)`

`=lim_(x->2)(2x^2-3x-2)/(x-2)`

`=lim_(x->2)((2x+1)(x-2))/(x-2)`

`=lim_(x->2) 2x+1`

`=5`