Question

Find the derivative using first principles? : `2sqrt(x)`

Answer 1

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

Explanation:

We have:

`f(x)=2sqrt(x)`

Then the limit definition of the derivative tells us that;

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

`\ \ \ \ \ \ \ \ \ = lim_(h rarr 0) ( 2sqrt(x+h)-2sqrt(x) ) / h`

`\ \ \ \ \ \ \ \ \ = 2lim_(h rarr 0) ( (sqrt(x+h)-sqrt(x)) ) / h * (sqrt(x+h)+sqrt(x))/(sqrt(x+h)+sqrt(x))`

`\ \ \ \ \ \ \ \ \ = 2lim_(h rarr 0) ( (x+h)-x ) / (h(sqrt(x+h)+sqrt(x))`

`\ \ \ \ \ \ \ \ \ = 2lim_(h rarr 0) ( (h ) / (h(sqrt(x+h)+sqrt(x))`

`\ \ \ \ \ \ \ \ \ = 2lim_(h rarr 0) 1 / (sqrt(x+h)+sqrt(x))`

Which we can evaluate, leading to the result:

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