Question

How do you find the derivative of `f(x) = sqrtx` using the formal definition?

Answer 1

I would use a "strange" rationalization of the numerator!

Explanation:

From the definition of derivative:
`lim_(h->0)((sqrt(x+h)-sqrt(x))/h)=`
I would try to rationalize the numerator (although it seems strange...).
`=lim_(h->0)((sqrt(x+h)-sqrt(x))/h)*color(red)(((sqrt(x+h)+sqrt(x)))/((sqrt(x+h)+sqrt(x))))=`
`=lim_(h->0)(x+h-x)/(h*(sqrt(x+h)+sqrt(x)))=`
`=lim_(h->0)(h)/(h*(sqrt(x+h)+sqrt(x)))=`
`=lim_(h->0)(1)/((sqrt(x+h)+sqrt(x)))=` as `h->0`
`=1/(2sqrt(x))`

Answer 2

Use: `f'(a) = lim_(h->0) ((f(a+h) - f(a))/h)`

to find `f'(x) = 1/(2sqrt(x))`

Explanation:

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

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

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

`=lim_(h->0) (((color(red)(cancel(color(black)(a)))+h)-color(red)(cancel(color(black)(a))))/(h(sqrt(a+h)+sqrt(a))))`

`=lim_(h->0) (1/(sqrt(a+h)+sqrt(a)))`

`=1/(2sqrt(a))`

That is:

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