What is the derivative of #log_2(x^2/(x-1))#?

Question

7480 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-21T03:18:32

#2/(xln(2))- 1/((x - 1)ln(2)) #

#log_2(x^2/(x - 1)) = #

#log_2(x^2)- log_2(x - 1) = #

#2log_2(x)- log_2(x - 1) = #

#(2/ln(2))ln(x)- 1/ln(2)ln(x - 1) #

Differentiate:

#(2/ln(2))1/(x)- 1/ln(2)1/(x - 1) = #

Simplify:

#2/(xln(2))- 1/((x - 1)ln(2)) #

Answer 2 · 2016-11-21T03:37:48

#=1/ln2((x-2)/(x(x-1)))#

Use #log_ba=lna/lnb#.

#(log_2(x^2/(x-1))'#

#=(ln (x^2/(x-1))/(ln2))'#

#=1/ln2(lnx^2-ln(x-1))'#

#1/ln2(2(lnx)'-(ln(x-1))')#

#=1/ln2(2/x-1/(x-1))#

#=1/ln2((x-2)/(x(x-1)))#