How do you find f'(x) using the definition of a derivative for $f (x) = \sqrt{2 x - 1}$ $f(x)=sqrt(2x-1)$ ?

Question

4264 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-05T20:11:37

See the explanation.

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

$f'(x)=lim_(h->0) (sqrt(2(x+h)-1)-sqrt(2x-1))/h=$

$=lim_(h->0) (sqrt(2(x+h)-1)-sqrt(2x-1))/h * (sqrt(2(x+h)+1)+sqrt(2x-1))/(sqrt(2(x+h)-1)+sqrt(2x-1))$

$=lim_(h->0) ((2(x+h)-1)-(2x-1))/(h(sqrt(2(x+h)-1)+sqrt(2x-1)))$

$=lim_(h->0) (2x+2h-1-2x+1)/(h(sqrt(2(x+h)-1)+sqrt(2x-1)))$

$=lim_(h->0) (2h)/(h(sqrt(2(x+h)-1)+sqrt(2x-1)))$

$=lim_(h->0) 2/(sqrt(2(x+h)-1)+sqrt(2x-1))$

$=lim_(h->0) 2/(2sqrt(2x-1))=1/sqrt(2x-1)$

How do you find f'(x) using the definition of a derivative for f(x)=sqrt(2x-1)f(x)=√2x−1f(x)=sqrt(2x-1)?