How do you differentiate #log_2(3x-1)#?

1 Answer
Apr 20, 2017

#3/((3x-1)*ln2)#

Explanation:

For this problem, the following formula will be used

#color(white)(aaaaaaaaaaa)##color(green)(d/dxlog_a[f(x)] = 1/(f(x)*lna) *f'(x)#

What this basically means, is that the derivative of a log base a of some function equals 1 over that function times the natural log of base a times the derivative of #f(x)#.

  • #d/dxlog_2(3x-1)#

  • #=1/((3x-1)*ln2)*3#

  • #=3/((3x-1)*ln2)#