How do you find the derivative of # log(8x-1)#?
1 Answer
Feb 9, 2016
#8/(8x - 1 )#
Explanation:
using the
#color(blue)(" chain rule ") #
# d/dx(f(g(x)) = f'(g(x) . g'(x) # and knowing that #d/dx(logx) = 1/x
# d/dx(log(8x-1)) = 1/(8x-1) d/dx (8x-1) = 8/(8x-1) #
