Question

How do you find the derivative of `sqrt(x-3)` using the limit process?

Answer 1

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

Explanation:

The limit definition of the derivative is

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

With `f(x) = sqrt(x-3)`, you can compute the derivative as follows:

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

... expand the fraction with `(sqrt(x+h-3) + sqrt(x - 3))`...

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

... apply the formula `(a+b)(a-b) = a^2 - b^2` to simplify the numerator...

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

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

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

... cancel `h`...

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

... at this point, you can safely apply the limit, i.e. plug `h = 0` in the denominator...

`= 1 / (sqrt(x+0-3) + sqrt(x - 3))`

`= 1 / (2 sqrt(x-3))`

Thus, your derivative is `1 / (2 sqrt(x-3))`.

Answer 2

(See below for process)
`f(x)=sqrt(x-3) rArr color(green)(f'(x)=1/(2sqrt(x-3))`

Explanation:

`f'(x) = lim_(hrarr0) (f(x+h)-f(x))/h`

`color(white)("XXX")=lim_(hrarr0)(sqrt(x+h-3)-sqrt(x-3))/h`

Multiply numerator and denominator by conjugate of denominator:
`color(white)("XXX")=lim_(hrarr0)((sqrt(x+h-3)-sqrt(x-3)))/h xx ((sqrt(x+h-3)+sqrt(x-3)))/((sqrt(x+h-3)+sqrt(x-3)))`

and since `(a-b)xx(a+b) = (a^2-b^2)`
`color(white)("XXX")=lim_(hrarr0)((x+h-3)-(x-3))/(h(sqrt(x+h-3)+sqrt(x-3)))`

`color(white)("XXX")=lim_(hrarr0)cancel(h)/(cancel(h)(sqrt(x+h-3)+sqrt((x-3)))`

`color(white)("XXX")=1/(sqrt(x-3)+sqrt(x-3))`

`color(white)("XXX")=1/(2sqrt(x-3)`