Question

How do you use the definition of a derivative to find the derivative of `f(x) = x + sqrtx`?

Answer 1

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

Explanation:

The definition of a derivative is

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

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

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

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

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

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

... expand the fraction so that you can use the formula `(a+b)(a-b) = a^2 - b^2` afterwards...

`= 1 + lim_(h->0) ((sqrt(x + h) - sqrt(x))* color(blue)((sqrt(x + h) + sqrt(x))))/(h * color(blue)((sqrt(x + h) + sqrt(x))))`

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

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

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

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

At this point, you can safely plug `h = 0` to compute the limit:

`= 1 + 1/(sqrt(x) + sqrt(x))`

`= 1 + 1/(2sqrt(x))`

Hope that this helped!