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

1 Answer
Oct 5, 2015

See the explanation.

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)