Using 1st principals solve √x³-x?

1 Answer
Jim H
Mar 5, 2018

Please see below.

Explanation:

We need the following facts from algebra:

#(sqrta-sqrtb)(sqrta+sqrtb) = a-b# and

#a^3-b^3 = (a-b)(a^2+ab+b^2)#

#f'(a) = lim_(xrarra)(f(x)-f(a))/(x-a)#

# =lim_(xrarra) (sqrt(x^3-x)-sqrt(a^3-a))/(x-a)#

# = lim_(xrarra)((sqrt(x^3-x)-sqrt(a^3-a)))/((x-a)) ((sqrt(x^3-x)+sqrt(a^3-a)))/ ((sqrt(x^3-x)+sqrt(a^3-a)))#

# = lim_(xrarra) ((x^3-x)-(a^3-a))/((x-a)(sqrt(x^3-x)+sqrt(a^3-a)))#

# = lim_(xrarra) ((x^3-a^3)-(x-a))/((x-a)(sqrt(x^3-x)+sqrt(a^3-a)))#

# = lim_(xrarra) ((x-a)(x^2+ax+a^2-1))/((x-a)(sqrt(x^3-x)+sqrt(a^3-a)))#

# = lim_(xrarra) (x^2+ax+a^2-1)/(sqrt(x^3-x)+sqrt(a^3-a))#

# = (a^2+aa+a^2-1)/(sqrt(a^3-a)+sqrt(a^3-a))#

# = (3a^2-1)/(2sqrt(a^3-a))#

Since #f'(a) = (3a^2-1)/(2sqrt(a^3-a))#, we have

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

Related questions
Impact of this question
854 views around the world
You can reuse this answer
Creative Commons License