How do you find f'(x) using the limit definition given $sqrt(2x-1)$ ?

Question

1557 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-14T21:36:46

$f'(x) = lim_{h to 0} (f(x+h) - f(x))/(h)$

$=lim_{h to 0} (sqrt(2(x+h)-1) - sqrt(2x-1))/(h)$

next, we times by the conjugate

$=lim_{h to 0} (sqrt(2(x+h)-1) - sqrt(2x-1))/(h) * (sqrt(2(x+h)-1) + sqrt(2x-1))/(sqrt(2(x+h)-1) + sqrt(2x-1))$

$= lim_{h to 0} (1/h) (2(x+h)-1 - (2x-1))/(sqrt(2(x+h)-1) + sqrt(2x-1))$

$= lim_{h to 0} (1/h) (2h)/(sqrt(2(x+h)-1) + sqrt(2x-1))$

$= lim_{h to 0} (2)/(sqrt(2(x+h)-1) + sqrt(2x-1))$

$= (2)/(sqrt(2x-1) + sqrt(2x-1))$

$= (2)/(2(sqrt(2x-1) )$

$= 1/(sqrt(2x-1)$

How do you find f'(x) using the limit definition given sqrt(2x-1)sqrt(2x-1)?