Question

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

Answer 1

`f'(x)=2x-3`

Explanation:

The definition of the derivative is:
`f'(x)=lim_(h->0)(f(x+h)-f(x))/h`

Applying this to the problem:
`f'(x)=lim_(h->0)((x+h)^2-3(x+h)-1-(x^2-3x-1))/h`
`f'(x)=lim_(h->0)(x^2+2xh+h^2-3x-3h-1-x^2+3x+1)/h`
`f'(x)=lim_(h->0)(2xh+h^2-3h)/h`
`f'(x)=lim_(h->0)(h(2x+h-3))/h`
`f'(x)=lim_(h->0)2x+h-3`
`f'(x)=2x+(0)-3=2x-3`

Answer 2

`(df(x))/dx=2x-3`

Explanation:

Definition of derivative is given by

`(df(x))/dx=Lt_(deltax->0)(f(x+deltax)-f(x))/(deltax)`

As `f(x)=x^2-3x-1`, hence

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

= `(x^2+2xdeltax+(deltax)^2-x^2)-3(x+deltax-x)-1+1`

= `(2xdeltax+(deltax)^2)-3(deltax)` and

`f((x+deltax)-f(x))/(deltax)=2x-3+deltax`

Hence, `Lt_(deltax->0)(f(x+deltax)-f(x))/(deltax)`

= `Lt_(deltax->0)(2x-3+deltax)` i.e.

`(df(x))/dx=2x-3`