How do you find f'(x) using the limit definition given $f (x) = \sqrt{x + 1}$ $f(x) =sqrt (x+1)$ ?

Question

1340 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-04T14:58:43

$= \frac{1}{2 \sqrt{x + 1}}$ $= 1/(2 sqrt (x+1) )$

$f (x) = \sqrt{x + 1}$ $f(x) =sqrt (x+1)$

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

use the conjugate

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

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

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

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

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

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

How do you find f'(x) using the limit definition given f(x) =sqrt (x+1)f(x)=√x+1 f(x) =sqrt (x+1)?