How do you find #f'(4)# if #f(x)=2sqrt(2x)#?

1 Answer
Mar 12, 2015

First find the derivative:
#f'(x)=2*2/(2sqrt(2x))=2/sqrt(2x)#
consider #sqrt(2x)# as #(2x)^(1/2)# and derive it as a normal power of #n# where if you have #f(x)=x^n# then #f'(x)=nx^(n-1)# and remember to derive also the argument #2x# (Chain Rule) that gives you simply #2#.
then substitute #x=4# in it:
#f'(4)=2/sqrt(8)=2/(2sqrt(2))=1/sqrt(2)#
rationalizing =#sqrt(2)/2#