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

2 Answers
Lotusbluete
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)).

Alan P.
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)

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