How do you find the derivative of #sqrt(x-3)# using the limit process?

2 Answers
Feb 5, 2016

#f'(x) = 1 / (2 sqrt(x-3)) #

Explanation:

The limit definition of the derivative is

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

With #f(x) = sqrt(x-3)#, you can compute the derivative as follows:

#f'(x) = lim_(h-> 0) (sqrt(x+h-3) - sqrt(x - 3))/h#

... expand the fraction with #(sqrt(x+h-3) + sqrt(x - 3))#...

# = lim_(h-> 0) ((sqrt(x+h-3) - sqrt(x - 3))(sqrt(x+h-3) + sqrt(x - 3)))/(h(sqrt(x+h-3) + sqrt(x - 3))) #

... apply the formula #(a+b)(a-b) = a^2 - b^2# to simplify the numerator...

# = lim_(h-> 0) ((sqrt(x+h-3))^2 - (sqrt(x-3))^2) / (h(sqrt(x+h-3) + sqrt(x - 3))) #

# = lim_(h-> 0) (x + h - 3 - (x- 3)) / (h(sqrt(x+h-3) + sqrt(x- 3))) #

# = lim_(h-> 0) " " h / (h(sqrt(x+h-3) + sqrt(x - 3))) #

... cancel #h#...

# = lim_(h-> 0) " " 1 / (sqrt(x+h-3) + sqrt(x - 3)) #

... at this point, you can safely apply the limit, i.e. plug #h = 0# in the denominator...

# = 1 / (sqrt(x+0-3) + sqrt(x - 3)) #

# = 1 / (2 sqrt(x-3)) #

Thus, your derivative is #1 / (2 sqrt(x-3))#.

Feb 5, 2016

(See below for process)
#f(x)=sqrt(x-3) rArr color(green)(f'(x)=1/(2sqrt(x-3))#

Explanation:

#f'(x) = lim_(hrarr0) (f(x+h)-f(x))/h#

#color(white)("XXX")=lim_(hrarr0)(sqrt(x+h-3)-sqrt(x-3))/h#

Multiply numerator and denominator by conjugate of denominator:
#color(white)("XXX")=lim_(hrarr0)((sqrt(x+h-3)-sqrt(x-3)))/h xx ((sqrt(x+h-3)+sqrt(x-3)))/((sqrt(x+h-3)+sqrt(x-3)))#

and since #(a-b)xx(a+b) = (a^2-b^2)#
#color(white)("XXX")=lim_(hrarr0)((x+h-3)-(x-3))/(h(sqrt(x+h-3)+sqrt(x-3)))#

#color(white)("XXX")=lim_(hrarr0)cancel(h)/(cancel(h)(sqrt(x+h-3)+sqrt((x-3)))#

#color(white)("XXX")=1/(sqrt(x-3)+sqrt(x-3))#

#color(white)("XXX")=1/(2sqrt(x-3)#