Use the definition of the derivative to find the derivative of `y=sqrt(2)`

By definition, the derivative of `f(x)` is `f'(x)=lim_(Delta x ->0)(f(x+Delta x)-f(x))/(Delta x)`

Thus:

`f'(x)=lim_(Delta x ->0)(sqrt(x+Delta x+2)-sqrt(x+2))/(Delta x)` "Rationalize" the numerator

`=lim_(Delta x ->0)((sqrt(x+Delta x +2)-sqrt(x+2)))/(Delta x)*((sqrt(x+Delta x+2)+sqrt(x+2)))/(sqrt(x+Deltax +2)+sqrt(x+2))` Do not distribute in the denominator

`=lim_(Delta x ->0)(x+Delta x+2-(x+2))/(Delta x(sqrt(x+Delta x+2)+sqrt(x+2)))`

`=lim_(Delta x ->0)(Delta x)/(Delta x(sqrt(x+Delta x+2)+sqrt(x+2)))` Since `Delta x !=0` we can cancel

`=lim_(Delta x ->0)1/(sqrt(x+Delta x+2)+sqrt(x+2))` Now we let `Delta x ->0`

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

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

**Thus the derivative is** `f'(x)=1/(2sqrt(x+2))`