How do you use first principles to show that if #f(x) = 3sqrt(x)#, then #f'(x) = 3/2x^(-1/2)#?

1 Answer
Noah G
Jul 2, 2017

#f'(x) = 3/2x^(-1/2)#. See below for details.

Explanation:

The first principle of differentiation involves the formula #f'(x) = lim_(h->0) (f(x + h) - f(x))/h#.

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

Let's try rationalizing the numerator to see what happens. This can be done by multiplying the entire fraction by #(3sqrt(x + h) + 3sqrt(x))/(3sqrt(x + h) + 3sqrt(x))#, this being because we don't want to change the value of the function.

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

#f'(x) = lim_(h->0) (9(x + h) - 9(x))/(3hsqrt(x + h) + 3hsqrt(x))#

#f'(x) = lim_(h->0) (9x + 9h - 9x)/(3hsqrt(x + h) + 3hsqrt(x))#

#f'(x) = lim_(h->0) (9h)/(3hsqrt(x+ h) + 3hsqrt(x))#

#f'(x) = lim_(h->0) (9h)/(h(3sqrt(x + h) + 3sqrt(x))#

#f'(x) = lim_(h->0) 9/(3sqrt(x + h) + 3sqrt(x))#

#f'(x) = 9/(3sqrt(x) + 3sqrt(x))#

#f'(x) = 9/(6sqrt(x))#

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

Hopefully this helps!

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