How do you find the derivative of #q(x) = (8x) ^(2/3)# using the chain rule?

1 Answer
Oct 31, 2015

Answer:

I found: #q'(x)=8/(3root3(x))#

Explanation:

First derive as a normal exponent (in red) and then multiply by the derivative of the argument (in blue):
#q'(x)=color(red)(2/3(8x)^(2/3-1))*color(blue)(8)=#
#=16/3(8x)^(-1/3)=#
#=16/3*1/root3(8x)=#
#=16/3*1/(2root3(x))=8/(3root3(x))#