What is the derivative of #1/sqrt(x-2)# using the limit definition? Sorry if I am asking an easy question!

1 Answer
Jul 22, 2015

This is not particularly easy using the limit definition, so that's fine. Using the limit definition:

#lim_(h->0) [f(x+h) - f(x)]/h#

Plug it in:
#= lim_(h->0) [1/(sqrt(x+h-2)) - 1/(sqrt(x-2))]/h#

Cross-multiply:
#= lim_(h->0) [sqrt(x-2)/(sqrt(x+h-2)sqrt(x-2)) - sqrt(x+h-2)/(sqrt(x-2)sqrt(x+h-2))]/h#

Combine:
#= lim_(h->0) ([sqrt(x-2) - sqrt(x+h-2)]/(sqrt(x+h-2)sqrt(x-2)))/h#

Move stuff around:
#= lim_(h->0) [sqrt(x-2) - sqrt(x+h-2)]/(h(sqrt(x+h-2)sqrt(x-2)))#

Multiply by the complex conjugate:
#= lim_(h->0) [sqrt(x-2) - sqrt(x+h-2)]/(h(sqrt(x+h-2)sqrt(x-2)))*(sqrt(x-2) + sqrt(x+h-2))/(sqrt(x-2) + sqrt(x+h-2))#

#= lim_(h->0) [x-2 - (x+h-2)]/(h(sqrt(x+h-2)sqrt(x-2))*(sqrt(x-2) + sqrt(x+h-2)))#

#= lim_(h->0) [-cancel(h)]/(cancel(h)(sqrt(x+h-2)sqrt(x-2))*(sqrt(x-2) + sqrt(x+h-2)))#

#= -lim_(h->0) 1/[(sqrt(x+h-2)sqrt(x-2))*(sqrt(x-2) + sqrt(x+h-2)))#

Hey, that looks nice now. You didn't even have to multiply out the denominator yet. So plugging in #h = 0#:

#= -lim_(h->0) 1/[(sqrt(x-2)sqrt(x-2))*(sqrt(x-2) + sqrt(x-2)))#

#= -lim_(h->0) 1/[(x-2)*(2sqrt(x-2)))#

#= color(blue)(-1/[2(x-2)^("3/2")])#

If you took the actual power rule derivative, you would get:

#d/(dx)[(x-2)^(-"1/2")] = -1/2*(x-2)^(-"3/2") = color(blue)(-1/[2(x-2)^("3/2")])#