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

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Answer 1 · 2014-10-17T02:09:13

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.