Question

Differentiate `f(x)=2x-3` using first principal?

Answer 1

`(df)/(dx)=2`

Explanation:

For a function `f(x)`, its derivative using first principal is given by

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

Here we have `f(x)=2x-3`,

hence `f(x+h)=2(x+h)-3=2x+2h-3`

and `f(x+h)=f(x)=2h`

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

= `Lt_(h->0)(2h)/h`

= `Lt_(h->0)2`

= `2`

Answer 2

`f'(x)=2`

Explanation:

`"differentiating from first principles"`

`f'(x)=lim_(hto0)(f(x+h)-f(x))/h`

`color(white)(f'(x))=lim_(hto0)(2(x+h)-3-2x+3)/h`

`color(white)(f'(x))=lim_(hto0)(2x+2h-2x)/h`

`color(white)(f'(x))=lim_(hto0)(2cancel(h))/cancel(h)=2`