How do you find the derivative of # f(x) = sqrtx# using the formal definition?

2 Answers
Oct 30, 2015

I would use a "strange" rationalization of the numerator!

Explanation:

From the definition of derivative:
#lim_(h->0)((sqrt(x+h)-sqrt(x))/h)=#
I would try to rationalize the numerator (although it seems strange...).
#=lim_(h->0)((sqrt(x+h)-sqrt(x))/h)*color(red)(((sqrt(x+h)+sqrt(x)))/((sqrt(x+h)+sqrt(x))))=#
#=lim_(h->0)(x+h-x)/(h*(sqrt(x+h)+sqrt(x)))=#
#=lim_(h->0)(h)/(h*(sqrt(x+h)+sqrt(x)))=#
#=lim_(h->0)(1)/((sqrt(x+h)+sqrt(x)))=# as #h->0#
#=1/(2sqrt(x))#

Oct 30, 2015

Use: #f'(a) = lim_(h->0) ((f(a+h) - f(a))/h)#

to find #f'(x) = 1/(2sqrt(x))#

Explanation:

#f'(a) = lim_(h->0) ((f(a+h) - f(a))/h)#

#=lim_(h->0) ((sqrt(a+h) - sqrt(a))/h)#

#=lim_(h->0) (((sqrt(a+h) - sqrt(a))(sqrt(a+h) + sqrt(a)))/(h(sqrt(a+h)+sqrt(a))))#

#=lim_(h->0) (((color(red)(cancel(color(black)(a)))+h)-color(red)(cancel(color(black)(a))))/(h(sqrt(a+h)+sqrt(a))))#

#=lim_(h->0) (1/(sqrt(a+h)+sqrt(a)))#

#=1/(2sqrt(a))#

That is:

#f'(x) = 1/(2sqrt(x))#