Question

How do you use the definition of a derivative to find f' given `f(x)=sqrt(4x+3)` at x>-3/4?

Answer 1

`f'=2/sqrt(4x+3)`

See below for derivation

Explanation:

The first principle defines the derivative of a function `f` as such:
`f'=lim_(h->0)(f(x+h)-f(x))/(h)`

We let `f=sqrt(4x+3)`
Then
`f'=lim_(h->0)(sqrt(4(x+h)+3)-sqrt(4x+3))/(h)`

`=lim_(h->0)1/h*(sqrt(4x+4h+3)-sqrt(4x+3))`

`=lim_(h->0)1/h*(sqrt(4x+4h+3)-sqrt(4x+3))*(sqrt(4x+4h+3)+sqrt(4x+3))/(sqrt(4x+4h+3)+sqrt(4x+3))`

`=lim_(h->0)1/h*((sqrt(4x+4h+3))^2-(sqrt(4x+3))^2)/(sqrt(4x+4h+3)+sqrt(4x+3))`

`=lim_(h->0)1/h*(4x+4h+3-4x-3)/(sqrt(4x+4h+3)+sqrt(4x+3))`

`=lim_(h->0)1/h*(4h)/(sqrt(4x+4h+3)+sqrt(4x+3))`

`=lim_(h->0)4/(sqrt(4x+4h+3)+sqrt(4x+3))`

`=4/(sqrt(4x+0+3)+sqrt(4x+3))`

`=4/(2sqrt(4x+3))`

`=2/sqrt(4x+3)`