How do you find the derivative of #y= root3(e^x+1)# ?
1 Answer
Sep 15, 2014
#y'=1/3(e^x+1)^(-2/3)*e^x# Explanation :
#y=(e^x+1)^(1/3)# let's
#y=(f(x))^(1/3)# , then from Chain Rule,
#y'=1/3(f(x))^(-2/3)f'(x)# Similarly following for the give problem and using Chain Rule, yields
#y'=1/3(e^x+1)^(-2/3)*e^x#
