How do I find the derivative of #f(x)=sqrt(x)# using first principles?

1 Answer
Oct 17, 2014

Definition

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


By Definition,

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

by multiplying the numerator and the denominator by #sqrt{x+h}+sqrt{x}#,

#=lim_{h to 0}{sqrt{x+h}-sqrt{x}}/hcdot{sqrt{x+h}+sqrt{x}}/{sqrt{x+h}+sqrt{x}}#

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

by cancelling out #x#'s and #h#'s,

#=lim_{h to 0}1/{sqrt{x+h}+sqrt{x}}=1/{sqrt{x+0}+sqrt{x}}=1/{2sqrt{x}}#

Hence, #f'(x)=1/{2sqrt{x}}#.


I hope that this was helpful.